Physics-20.13-Electrostatics
13-Electrostatics
ELECTROSTATICS
ELECTRIC CHARGE
Charge is the physical property of certain fundamental particles (like electron, proton). Charges are divided into two parts (i) positive (ii) negative. Like charges repel and unlike charges attract. SI unit of charge is coulomb and CGS unit is esu. 1C = 3 109esu.
Magnitude of the smallest known charge is e = 1.6 10-19C (charge of one electron or proton).
PROPERTIES OF ELECTRIC CHARGE
QUANTIZATION OF CHARGE
charge on any body is the integral multiple of the charge on an electron
⇒ Q = ne, where n = 0, 1, 2, .................
Distribution of Charges
The concentration of the charges is maximum on a surface with greater curvature.
COULOMB’S LAW
The force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them, In free space
F = ∈- permittivity of the medium
In material medium
F = ∈0 – permittivity of free space (Vacuum)
Where ∈ = ∈0 ∈r ∈r = relative permittivity of the medium
in vector form
∈0 = 8.85410-12
1. This is a fundamental law and is based on physical observation
2. The force is an action - reaction pair.
3. The direction of force is always along the line joining the two charges.
4. Electrical force between two point charges is independent of presence or absence of other charges.
Example 1. A particle ‘A’ having a charge of 2 10-6C and a mass of 100g is fixed at the bottom of a smooth inclined plane of inclination 30. Where should another particle B, having same charge and mass be placed on the incline so that it may remain in equilibrium ?
Solution : First of all draw the F.B.D. of the masses.
For equilibrium F = 0
N = mg cos30 . . . (1)
Fe = mg sin30 . . . (2)
From (2)
⇒ x = = 27cm.
Electric Field Intensity It is the force experienced by a unit positive charge placed at a point in an electric field.
[In vacuum or free space]
N/coul or Volt/m
Example 2.: Two particles A and B having charges 8 x10-6 C and –2 x10-6C respectively are held fixed with a separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force ?
Solution : As the net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have charges of opposite signs, C cannot be between A and B
Also A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. The situation is shown in figure. Suppose BC=x and the charge on C is Q
and
But
Hence
Which gives x = 0.2 m
Superposition principle
Force experienced by a given charge in the field of a number of point charges is the vector sum of all the forces.
Similarly for electric field
∴ Lines of Force
∙ Lines of force originate from positive charges and terminate on negative charges
∙ Lines of force originate or terminate perpendicular to the surface
∙ Tangent to the lines of force at any point gives the direction of the electric field.
∙ Lines of force do not intersect.
1. Electric Field Intensity
Due to a point charge
E =
Due to linear distribution of charge
= linear charge density of rod
(i) At a point on its axis.
E = = linear charge density
(ii) At a point on the line
perpendicular to one end
Ex =
Ey =
Due to ring of uniform charge distribution
At a point on its axis
Due to uniformly charged disc.
At a point on its axis.
= surface charge density
Thin spherical shell
2. Non conducting solid sphere having uniform volume distribution of charge
(i) Outside point
E =
(ii) Inside point q = total charge
= volume density of charge
Cylindrical Conductor of Infinite length
Outside point
Inside point = linear density of charge
E = 0 (as charges reside only on the surface)
Non-conducting cylinder having uniform volume density of charge
Outside point
Inside point
Infinite plane sheet of charge
E = = surface charge density
Two oppositely charged sheets (Infinite)
(i) Electric field at points outside the charged sheets
EP = ER = 0
(ii) Electric field at point in between the charged sheets
EQ =
Example 3. Find electric field intensity due to long uniformly charged wire.
(charge per unit length is )
Solution: Elemental charge dq = d.
Field at point P is dE =
Also
3. On integration
4.
Example 4. What is the electric field at any point on the axis of a charged rod of length `L' and linear charge density`'? The point is separated from the nearer end by a.
Solution : Consider an element, dx at a distance x from the point, P, where we seek to find the electric field. The elemental charge, dq =dx
[ k = ]
Example 5. A ring shaped conductor with radius R carries a total charge q uniformly distributed around it as shown in the figure. Find the electric field at a point P that lies on the axis of the ring at a distance x from its centre.
Solution: Consider a differential element of the ring of length ds. Charge on this element is dq = .
This element sets up a differential electric field d at point P.
The resultant field at P is found by integrating the effects of all the elements that make up the ring. From symmetry this resultant field must lie along the right axis. Thus, only the component of d parallel to this axis contributes to the final result.
= ⇒ = cos
dE =
cos =
To find the total x-component Ex of the field at P, we integrate this expression over all segment of the ring.
Ex = cos =
The integral is simply the circumference of the ring = 2R
E =
As q is positive charge, field is directed away from the centre of the ring, along its axis.
5. DIPOLE
Two equal and opposite charges separated by a small distance constitute an electric dipole
is directed from –ve to the +ve charge
(i) Electric field at an Axial Point
E =
For x >> a
.
(ii) At a point on its perpendicular bisector
E =
For y >> a
(iii) At any point
r >> a
and
∴
Angle = tan-1 (1/2 tan )
6. GAUSS THEOREM
The flux of an electric field through an arbitrary closed surface S is equal to the algebraic sum of the charge enclosed by this surface divided by ∈0.
Example 6. Figure shows a section of an infinite rod of charge having linear charge density which is constant for all points on the line. Find electric field E at a distance r from the line.
Solution : From symmetry, due to a uniform linear charge can only be radially directed. As a Gaussian surface, we can choose a circular cylinder of radius r and length l, closed at each end by plane caps normal to the axis.
o = qin
Cylindrical Plane Surface
o E (2rl) +
E =
The direction of is radially outward for a line of positive charge.
Example 7. Figure shows a sphericaly symmetric distribution of charge of radius R. Find electric field E for points A and B which are lying outside and inside the charge distribution respectively.
Solution: The spherically symmetric distribution of charge means that the charge density at any point depends only on the distance of the point from the centre and not on the direction. Secondly, the object can not be a conductor, or else the excess charge will reside on its surface.
Now, apply Gauss’s law to a spherical Gaussian surface of radius r ( r > R for
point A)
o ⇒ = q
E = where q is the total charge
For point B (r> a
V =
8. Potential due to a uniformly charged disc
(i) At a point on its axis
V = = surface charge density on disc
(ii) At the centre of the disc
V =
9. Potential at the edge of a uniformly charged disc
V =
Potential at the apex of a cone having charge Q distributed uniformly on its curved surface and having slanting length L
V =
∴ Potential of a conducting sphere of radius R at a distance r from the centre
(i) r > R V = q = charge on the sphere
(ii) r = R V = K =
(iii) r < R
10. Potential of a dielectric sphere
(i) r > R q – total charge in the dielectric sphere
V =
(ii) r = R V =
(iii) r < R V =
Torque on a dipole in presence of external electric field
Torque = qE. 2a sin
= PE sin
Potential Energy U =
Example 8. Two conducting spheres having radii a and b charged to q1 & q2 respectively. Find the potential difference between 1 & 2 ?
Solution : The potential on the surface of the sphere 1 is given by
v1 = . . . . (A)
The potential on the surface of the sphere 2 is given by,
v2 = ,
⇒
⇒
11. Electric Potential Energy
The electric potential energy of a system of point charges is the amount of work done in bringing the charges from infinity in order to form the system. For point charges q1 and q2 separated by a distance r12, Electric potential energy of the system q1 and q2 is given by
U =
For three particle system q1, q2 and q3
U =
We can define electric potential (VP) at any point P in a electric field as , VP = ; where Up, is the change in electric potential energy in bringing the test charge q0 from infinity to point P.
Example 9. Determine the interaction energy of the point charges of the following set- up
Solution : U = U12 + U13 + U14 + U23 + U24 + U34
= − 3 (Here k = 1/ 40)
12. CAPACITORS
13. Capacitance
If Q is the charge given to a conductor and V is the potential to which it is raised by this amount of charge, then it is found that Q ∝ V. Or Q = CV, where C is a constant called capacitance of the conductor.
C =
For parallel plate capacitor C =
i.e. the capacitance depends only on geometrical factors namely, the plate area and plate separation.
∴
14. Spherical Capacitor
C =
15. Isolated Capacitor
An isolated sphere can be thought of as a capacitor where other plate is at infinity.
R1 = R and R2 = ∞
∴ C =
16. Cylindrical Capacitor
C =
17. 18.
∙ Combination of Capacitors
Series Combination: In series combination, each capacitors has equal charge for any value of capacitance. Equivalent capacitance C is given by
Parallel Combination:
In parallel combination the potential differences of the capacitor connected in parallel are equal for any of capacitor. Equivalent capacitance is given by
C = C1 + C2 + C3
Example 10. In the above circuit, find the potential difference across AB.
Solution: Let us mark the capacitors as 1, 2, 3 and 4 for identifications. As is clear, 3 and 4 are in series, and they are in parallel with 2. Then 2, 3, 4 combine is in series with 1.
The `q’ on 1 is 48 C, thus V1=q/c=6v [v1 = ]
⇒ VPQ = 10 − 6 = 4V
By symmetry of 3 and 4, we say, VAB = 2V.
Dielectric
Substances having polar atoms/molecules intrinsically or being polarized are called dielectrics.
19. Polar Dielectric
Substances which have polar atoms/molecules intrinsically but are randomly arranged. On application of external electric field they get polarized parallel to the external electric field.
Net electric field inside dielectric.
= electric field due to induced charges
– surface charge density of capacitor plates
i – induced charge density
20. Capacitance of a parallel plate capacitor having dielectric slabs in series.
C =
21. Energy stored in a Capacitor
Energy supplied by Source
US = Q. V.
Energy lost in form of heat =
22. Energy density of the electric field
in free space
in presence of dielectric medium ; K - dielectric const.
23. Force on a Dielectric in a Capacitor
when source is connected
V =
⇒
Example 11. A 4f capacitor is charged to 150 V and another 6f capacitor is charged to 200 V. Then they are connected across each other. Find the potential difference across them. Calculate the heat produced.
Solution : 4f charged to 150 V would have q1 = C1V1 = 600C
6f charged to 200 V would have q2 = C2V2 = 1200C
After connecting them across each other, they will have a common potential difference V.
Charges will readjust as q1’ and q2’
Final energy
= 0.161J
Heat produced = |Uf − Ui| = 0.003 J
OBJECTIVE
1 : A long string with a charge of per unit length passes through an imaginary cube of edge a. The maximum flux of the electric field through the cube will be
(A) (B)
(C) (D)
Solution : The maximum length of the string which can fit into the cube is a, equal to its body diagonal. The total charge inside the cube is , and hence the total flux through the cube is .
∴ (D)
2 : Potential in the x-y plane is given as V = 5(x2 + xy) volts. The electric field at the point (1, −2) will be
(A) 3J V/m (A) −5 J V/m
(C) 5J V/m (D) −3J V/m
Solution : Ex = − = −(10 x + 5y) = −10 + 10 = 0
Ey = − = −5x = −5
∴ .
∴ (B)
3 : In the circuit shown, the equivalent capacitance between the points A and B is
(A) C/5 (B) C/3
(C) C/2 (D) C
Solution: Rearranging the circuit, the points E and D are at the same potentital (by symmetry). Then the capacity between E and D can be removed.
∴ C′ = C/2
and are in parallel.
Hence Ceq = = C
∴ (D)
4 : Electric field in a region is given by V/m. The potential difference between points (0, 0, 0) and (1,2,3) will be
(A) 2 V (B) 5 V
(C) 4V (D) 6 V
Solution: P.d. across the points = −
V2 – V1 = − = −2 − 6 + 12 = 4 volts.
∴ (C)
5: What is the mechanical work done in pulling the slab out of the capacitor after disconnecting it from the battery
(A) (B) E2C
(C) E2C (r − 1) (D) none of these
Solution: Work done = change in potential energy = U2 – U1
v1 = (1/2) E2C
v2 = = (1/2) E2 Cr
∴ Work done = (1/2) E2C (r – 1).
∴ (A).
6 : Two point charges each of charge +q are fixed at
(+a, 0) and (-a, 0). Another positive point charge q placed at the origin is free to move along x- axis. The charge q at origin in equilibrium will have
(A) maximum force and minimum potential energy.
(B) minimum force & maximum potential energy.
(C) maximum force & maximum potential energy.
(D) minimum force & minimum potential energy.
Solution: The net force on q at origin is
=
The P.E. of the charge q in between the extreme charges at a distance x from the origin along +ve x axis is
U =
= .
For U to be minimum
⇒ (a-x)2 = (a + x)2
⇒ a + x = (a – x)
⇒ x = 0, because other solution is irrelevent.
Thus the charged particle at the origin will have minimum force and minimum P.E.
∴ (D).
7 : In a parallel plate capacitor of capacitance C, a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plates. The capacitance now becomes
(A) 4C (B) 2C
(C) C/2 (D) C/4
Solution : Before the metal sheet is inserted, C =
After the sheet is inserted, the system is equivalent to two capacitors in series, each of capacitance
The equivalent capacity is now 2C.
∴ (B)
8 : The value of q if it floats in air is
(A) (B)
(C) (D) none of these
Solution : The magnitude of electric field due to the charged conducting plate is E =
As the charged particle is floating in air (neglecting the buoyant force due to air we obtain)
mg = qE
⇒ q =
⇒ q =
∴ (A)
9 : The capacitance of an filled parallel plate capacitor is 20F. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of 4010−12 farads. The dielectric constant of wax is
(A) 12.0 (B) 10.0
(C) 8.0 (D) 4.2
Solution : 1010−12 = . 4010−12 =
∴ K = 8
∴ (C)
10 : The ratio of the time periods of small oscillation of the insulated spring and mass system before and after charging the masses is
(A) ≥ 1 (B) > 1
(C) ≤ 1 (D) = 1
Solution : In equilibrium of the charged small bodies = kx0 where x0 is the elongration in the spring in equilibrium.
Let a further small elongation of x is made in the spring.
Then net restoring force on any of the charged particle is given by,
F = −
= −kx. Since x < < x0 from (1)
⇒ a = − x
As F = a where = ⇒ a = − 2 x,
Hence = ⇒ T = 2
In absence of charge is T0 = 2 .
Therefore
∴ (D)
11: The charge flowing across the cell on closing the key k is equal to
(A) CV (B) CV/2
(C) 2CV (D) zero
Solution: When the key is kept open, the charge drawn from the source is
Q = CeqV = V
When the key is closed the capacitor 2 gets short circuited
And C′eq = C
∴ Q′ = CV
charge flown through cell Q′− Q = V
∴ (B) is correct choice.
12: The conducting spherical shells shown in the figure are connected by a conductor. The capacitance of the system is
(A) 40 (B) 40 a
(C) 40 b (D) 40
Solution: Hence, the capacitance of the system is the capacitance due to outer sphere of radius b.
∴ C = 4ob
(C) is correct choice.
13: A point charge q moves from point P to S along the path PQRS in a uniform electric field pointing parallel to the positive direction of the x-axis. The coordinate of the point P, Q, R and S are (a, b, 0), (2a, 0, 0), (a, −b, 0) and
(0, 0, 0) respectively. The work done by the field in the above process is given by the expression
(A) qaE (B) − qaE
(C) q E (D) 3qE
Solution: The work done is independent of the path followed and is equal to ,
where = displacement from P to S
Here, = , while = E
∴ work = −(qE ). = − qaE
Hence, B is correct choice.
14: Charge Q is given to the upper plate of an isolated parallel plate capacitor of capacitance C. The potential difference between the plates
(A) (B)
(C) (D) zero
Solution: In general, for charge Q1 and Q2on upper and lower plate respectively the charge distributions on outer and inner part of the plates are shown in figure.
Here Q1 =Q, Q2 = 0
∴ Charge on inner side of plate are and − respectively.
Hence V =
Hence (C) is correct choice.
15: A particle A of mass m and charge Q moves directly towards a fixed particle B. Which has charge Q. The speed of A is v when it is far away from B. The minimum separation between the particle is proportional to
(A) v (B) v2
(C) v−1 (D) v−2
Solution: From Conservation of energy
(KE+EPE)minimum separation = (KE+EPE)far away
⇒ 0 +
⇒ rmin ∝
Hence, (D) is correct choice.
16: A dipole of dipole moment is kept along an electric field such that and are in the same direction. Find the work done in rotating the dipole by an angle .
(A) W = 2Ep (B) W = –2Ep
(C) W = Ep (D) W = –Ep
Ans. (A)
Solution : W = U = U2 – U1
Now U2 = −(Ep cos ) = Ep
U1 = −(Ep cos 0) = −Ep
∴ W = 2Ep.
17: A simple pendulum of mass m and length carries a charge q. Find its time period when it is suspended in a uniform electric field region as shown in figure.
(A) T = 4 (B) T = 2
(C) T = 2 2 (D) T = 2
Ans. (B)
Solution: Time period of the pendulum
= 2
Here, geff =
=
=
∴ T = 2 .
18: A particle of mass 100 gm and charge 2 C is released from a distance of
50 cm from a fixed charge of 5C. Find the speed of the particle when its distance from the fixed charge becomes 3 m. Neglect any other force.
(A) –1.73 m/s (B) 2.73 m/s
(C) –2.73 m/s (D) 1.73 m/s.
Ans. (D)
Solution: Loss of potential energy = gain in kinetic energy
U1 – U2 = K.
KQq = ½ mv2 – 0
= = 1.73 m/s.
19: If a point charge q is placed at the centre of a cube what is the flux linked
with the cube?
(A) (B)
(C) (D)
Ans. (A)
Solution : From gauss’s law, flux linked with a closed body is times the charge enclosed. The cube encloses a charge q so flux linked with cube,
20: If a point charge q is placed at the centre of a cube what is the flux linked
with each face of the cube?
(A) (B)
(C) (D)
Ans. (B)
Now as cube is a symmetrical body with a faces and the point charge is at its centre, so electric flux linked with each face will be
21: Supposing that the earth has a surface charge density of 1 electron/m2; calculate
earth's potential
(A) − 0.115Volt (B) − 0.110 Volt
(C) − 0.105Volt (D) − 0.112 Volt
Ans. (A)
Solution: If R be the radius and the surface charge density of earth, then the total charge q on its surface is given by
Q = 4R2
(i) The potential V at a point on earth's surface is same as if the entire charge q were concentrated as its centre. Thus
V = =
Substituting the given value:
V =
= − 0.115 N-m/C = −0.115 J/C = − 0.115Volt.
22: Supposing that the earth has a surface charge density of 1 electron/m2; calculate
electric field just outside earth's surface.
The electronic charge is − 1.610−19C and earth's radius is 6.4106m.
(0 = 8.910−12C2/N-m2)
(A) − 1.810−8 N/C (B) + 1.810−8 N/C
(C) − 1.810−9 N/C (D) + 1.810−9 N/C
Ans. (A)
Solution: (ii) Again, the electric field E just outside the earth's surface is same is if the entire charge q were concentrated at this cetnre. Thus
E = =
Substituting the given value:
E =
= − 1.810−8 N/C
The minus sign indicates that E is radially inward.
23 : A circular wire of radius R carries a total charge Q distributed uniformly over its circumference. A small length of the wire subtending angle at the centre is cut off. Find the electric field at the centre due to the remaining portion .
(A) (B)
(C) (D)
Ans. (B)
Solution : Electric field due to an arc at its centre is , Where k = ,
= angle subtended by the wire at the centre,
= Linear density of charge.
Let E be the electric field due to remaining portion.
Since intensity at the centre due to the circular wire is zero.
Applying principle of superposition.
24: An electric dipole, made up of a positive and a negative charge, each of 1C and placed at a distance 2 cm apart, is placed in an electric field 105N/C. Compute the maximum torque which the field can exert on the dipole, and the work that must be done to turn the dipole from a position = 0 to = 180
(A) 410−6N-m or Joule (B) 410−9N-m or Joule
(C) 410+6N-m or Joule (D) 410−3N-m or Joule
Ans. (D)
Solution : The torque exerted by an electric field E on a dipole of moment p is given by
= pE sin,
Where is the angle which the dipole is making with the field.
is a maximum, when = 90. That is
∴ max = pE
Here p = q(2) = 110−6 0.02C/m and E = 105N/C
∴ max = 110−6 0.02 105 = 210−3N-m
The work done in rotating the dipole from an angle o to is given by
W = = pE (coso− cos)
Here o = 0 and = 180
∴ W = pE (cos0− cos180) = 2pE = 410−3N-m or Joule
25: A parallel-plate air capacitor has a plate area of 100cm3 and separation 5mm. A potential difference of 300V is established between its plates by a battery. After disconnecting a battery, the space between the plates is filled by ebonite (K = 2.6). Find out new potential difference between plates of the capacitor.
(A) 115V (B) 120V
(C) 110V (D) 125V
Ans. (A)
26: A parallel-plate air capacitor has a plate area of 100cm3 and separation 5mm. A potential difference of 300V is established between its plates by a battery. After disconnecting a battery, the space between the plates is filled by ebonite (K = 2.6). Find out initial and final capacitances of the capacitor.
(A) 2.7710−11F, 4.610−11F (B) 1.7710−11F, 4.610−11F
(C) 1.7710−11F, 6.610−11F (D) 1.7710−13F, 4.610−11F
Ans. (B)
27: A parallel-plate air capacitor has a plate area of 100cm3 and separation 5mm. A potential difference of 300V is established between its plates by a battery. After disconnecting a battery, the space between the plates is filled by ebonite (K = 2.6). Find out initial and final surface-density of charge on the plates.
(A) 5.3110+7C/m2 (B) 6.3110−7C/m2
(C) 5.3110−7C/m2 (D) 9.3110−7C/m2
Ans. (C)
Solution: Capacity of the parallel plate air capacitor
= = 1.7710−11F
Final capacity of the capacitor with dielectric between the plates is
C′ = KC = 2.61.7710−11, C′ = 4.610−11F
Initial charge on the capacitor 1.7710−11 300 = 5.3110−9C
Since, the battery has been disconnected, the charge remains the same, therefore the new potential difference is
V′ = = 115V
The surface density of charge remains the same in both the cases, i.e.,
= = = 5.3110−7C/m2
28: A spherical condenser has 10cm and 12 cm as the radii of inner and outer spheres. The space between the two is filled with a dielectric of dielectric constant 3. Find the capacity when
the outer sphere is earthed.
(A) 410−10F (B) 110−10F
(C) 210−10F (D) 610−10F
Ans. (C)
Solution: C = 4k0 = = 210−10F
29: A spherical condenser has 10cm and 12 cm as the radii of inner and outer spheres. The space between the two is filled with a dielectric of dielectric constant 3. Find the capacity when
the inner sphere is earthed.
(A) (B)
(C) (D)
Ans. (D)
By choice of reference potential at infinity and the potential of earthed conductor zero, the given system can be visualised as combination of two spherical capacitors, both being at same potential difference. Connection wise these may be considered to be in parallel connection.
∴ C = 4ob + 4k0 =
30: A charge Q is distributed over two concentric hollow spheres of radii r and R (R >r) such that the surface densities are equal. Find the potential at the common centre.
(A) (B)
(C) (D)
Ans. (A)
Solution: q1 + q2 = Q . . . (i)
= . . . (ii)
from (i) and(ii)
q1 = q2 =
Vcentre = V1 + V2 =
=
31: An electric dipole of dipole moment P is placed in a uniform electric field E is stable equilibrium position. Its moment of inertia about the centroidal axis is I. If it is displaced slightly from its mean position find the period of small oscillation.
(A) (B)
(C) (D)
Ans. (B)
Solution: When displaced at an angle , from its mean position the mean position the magnitude of restoring torque is
For small angular displacement sin ≈
The angular acceleration is,
Where
∴
32: Figure shows two conducting thin concentric shells of radii r and 3r. The outer shell carries charge q. Inner shell is neutral. Find the charge that will flow from inner shell to earth after the switch S is closed.
(A) (B)
(C) (D)
Ans. (D)
Solution: Let q′ be the charge on inner shell when it is earthed.
Vinner = 0
∴
∴
i.e. charge will flow froms inner shell to earth.
33: A capacitor stores 10C charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 20C flows through the battery. Find the dielectric constant of the dielectric.
(A) k = 2 (B) k = 4
(C) k = 3 (D) k = 1
Ans. (C)
Solution: In absence of dielectric
Q = CV = 10C . . . (1)
With dielectric
Q′ = kCV = 30C . . . (2)
From (i) and (2)
k = 3
34: Two infinitely large sheets S1 and S2 having surface charge densities 1 and 2 (1 > 2) respectively are placed at a distance d. Find the work done by the electric field when a charge particle ‘ Q’ is displaced by a distance ‘a’ (a < d) at an angle 450 with the normal of the sheets. It is assume that the charge does not affect the surface charge densities of the two plates.
(A) (B)
(C) (D)
Ans. (C)
Ans. Electric field at any point between the plate
=
35.: A positive charge (+q) is located at the centre of a circle as shown in figure. W1 is the work done in taking a unit positive charge from A to B and W2 is the work done in taking the same charge from A to C. Then.
(A) W1 > W2 (B) W1 < W2
(C) W1 = W2 (D) W1 = W2 = 0
Solution: Point A, B and C are at the same distance from charge +q; hence electrical potential is the same at these points, i.e. There is no potential difference between A, B and C. Hence W1 = W2 = 0.
36. A point Charge q is placed inside the cavity of a metallic shell.
Which one of the diagram correctly represents the electric lines of force.
(A)
(B)
(C)
(D)
Solution: (C) Because electric field inside the conductor is zero and electric field lines are perpendicular to Gaussian surface.
37.: A soap bubble has radius R, charge Q, surface tension T. Find the excess pressure in it.
(A) (A)
(C) (D) none of these
Solution: (C)
38.. A charge of 2C is brought from B to C along the path as shown by arrow in the figure. The work done is
(A) 0.75 J (B) 0.6J
(C) 0.06J (D) 0.075J
Solution: (D)
= 0.075 J
39. The electric field intensity at a point is . Considering potential at origin to be zero, the potential at P (2, 2) is
(A) (B)
(C) –100V (D) 20V
Solution: (C)
40. The equivalent capacitance between A and B is (each of the capacitors obtained is of capacitance equal to C)
(A) (B)
(C) (D)
Ans. (C)
41. 28. Page 426 Four large metal plates are located a small distance apart from one another as shown in the Fig. 24. The extreme plates are connected by means of a conductor, while a potential difference V is applied to the internal plates. Find the electric fields between the neighbouring plates
(A) (B)
(C) zero (D) None of these
Ans. (A) We may take the plates to be connected as shown in the Fig. 56. Let E23 be the electric field between plates 2 and 3. Then from the fact that electric field = rate of change of potential, .
Applying `loop rule' to the loop
`A – 1 – 2 – 3 – 4 – B – A'
V' – V – + V' ⇒ V' =
where V' is the numerical value of the potential difference
between plates 1 and 2 or 3 and 4
(density of charge on the plate 1)
(numerically)
The surface density of charge of the right surface of plate
2 is
(surface density of charge on plate 2)
Thus
(i)
(ii) .
42. A simple pendulum consists of a small sphere of mass m suspended by a thread of length l. The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength E directed vertically upwards. With what period will the pendulum oscillate if the electrostatic force acting on the sphere is less that the gravitational force? Assume the oscillations to be small
(A) (B)
(C) (D)
Solution: Let x be the small displacement given to the pendulum such that the angle is small. The forces acting at A are
(i) tension T along the thread
(ii) weight mg acting vertically downwards.
(iii) electrical force qE vertically upwards.
The resultant force vertically down wards is (mg – qE). Therefore
Net acceleration
Time period
43. A ring has charge Q and radius R. If a charge q is placed at its centre then the increase in tension in the ring is
(A) (B) zero
(C) (D)
Solution: Consider a small element AB, is very small. Then
AB = R(2)
Change on AB is
44. A charge Q is uniformly distributed over a large plastic plate. The electric field at point P, close to the centre of the plate is 10Vm-1 . If the plastic plate is replaced by a copper plate of the same dimensions and carrying same charge Q then the electric field at the point P will be
(A) 5Vm-1 (B) zero
(C) 10Vm-1 (D) 20Vm-1
Solution: (C) because E =
45. A hemisphere of radius r is placed in a uniform electric field of strength E. The electric flux through the hemisphere is
(A 2Er2 (B) –Er2
(C) -2Er2 (D) zero
Solution:
short cut AXB is symmetrical surface
Electric flux due to this part is zero. However, electric flux due to AB part is –ER2.
46. 1. What is the potential difference at the centre C
(A) Zero (B)
(C) (D) none of these
Solution: (A) We know that potential is a scalar quantity
V =
Where r is the distance between charge and point where potential has to be found.
∴ total potential at center C
= (q1 + q2 + q3 + q4).
= (+ q – q + q – q)
= 0
47. MeV is
(A) (B)
(C) (D)
Solution: (C) 1 MeV = 1.6 10–16 106 Joules
= 1.6 10–13 Joules
48. What is the angle between electric dipole moment and the electric field strength due to it on the axial line
(A) 0 (B) 90
(C) 180 (D) none of these
Solution: (D) Direction of dipole movement will be opposite to the electric field
49. The total electric flux, leaving spherical surface of radius 1 cm, and surrounding an electric dipole is
(A) (B) zero
(C) (D)
Solution: Electric flux passes as due to q charge of dipole
Electric flux passes due to – q charge of dipole
Net flux passes due to both the charges 0
50. An isolated sphere of radius R contains a uniform volume distribution of positive charge. Which of the curve on the graph below correctly illustrates the dependence of the magnitude of the electric field of the sphere as a function of the distance r from its centre
(A) A (B) B
(C) C (D) D
Solution: (C) E r for inside point and E for outside point and it
will be represented by C.
51. The variation of potential with distance R from a fixed point is as shown in figure
(A) the electric field at
(B) the electric field at ,
(C) the potential at R = 5m is + 2.5 V- m– 1 is continuous and a
decreasing function of distance.
(D) the segment AB corresponds to an equipotent surface.
Solution: (D) (The segment AB corresponding to an equipotential surface)
52. An electric dipole is placed inside a conducting shell. Mark the correct statement(s)
(A) the flux of the electric field through the shell is zero
(B) the electric field is zero at every point on the shell
(C) the electric field is not zero anywhere any where on the shell
(D) the electric field is zero on a circle on the shell.
Solution: (A) Inside the surface the total charge is zero. So flux must be zero.
53. A network of six identical capacitors, each of value C is made as shown in the figure. Equivalent capacitance between points A and B is
(A) C/4 (B) 3C/4
(C) 4C/3 (D) 3C
Solution: (C) The network is equivalent to –
Therefore equivalent capacitance = [2C series C] // [C series 2C]
54. The charge flowing across the cell on closing the key k is equal to
(A) CV (B) CV/2
(C) 2CV (D) zero
Solution: When the key is kept open, the charge drawn from the source is
Q = CeqV = V
When the key is closed the capacitor 2 gets short circuited
And C′eq = C
∴ Q′ = CV
charge flown through cell Q′− Q = V
∴ (B) is correct choice.
55. The capacitance of the system of parallel plate capacitor shown in the figure is
(A) (B)
(C) (D)
Solution : Since the electric field between the parallel charge plates is uniform and independent of the distance, neglecting the edge effect, the effective area of the plate of area A2 is A1. Thus the capacitance between the plates is
C =
∴ (C)
56. The three capacitors in figure, store a total energy in J of (A) 12 (B) 36
(C) 48 (D) 80
Solution: (C) Total capacitance 6 F
Applied voltage 4v
Stored energy U
57. In the circuit shown in figure C = 6 F. The charge stored in capacitor of capacity C is
(A) zero (B) 90 C
(C) 40 C (D) 60 C
Solution: (C) Both the capacitors are in series. Therefore charge stored on them
will be same.
Net capacity
Potential difference 10V
∴ q CV 40C
58. Figure shows two parallel plates, R and S joined to a battery of voltage V and with charges +Q and –Q.
1. the energy stored is QV.
2. the electric field strength between the plates increases uniformly from S to R.
3. the electric potential between the plates decreases uniformly from R to S. Which of the statement is/are correct
(A) 3 only (B) 1 and 2 only
(C) 2 and 3 only (D) 1 only
Solution: (A) The electric potential between the plates decreases uniformly
from R to S.
59. In the circuit shown in figure potential difference between A and B
is
(A) 30 V (B) 60 V
(C) 10 V (D) 90 V
Solution: (C) Potential difference is divided among two capacitance C1 and C2 in the inverse ratio of their capacities when it is final in series. Therefore
Voltage across P and Q is
equivalent capacitance between P and Q
Potential difference between P and Q (i.e. 40 volt)
Will divide between two capacitor C and 3C which is in series. Therefore voltage across 3C capacitance.
Hence (C) is correct choice
60. A parallel plate capacitor has two layers of dielectrics as shown in figure. This capacitor is connected across a battery, then the ratio of potential difference across the dielectric layers is
(A) 4/3 (B) 1/2
(C) 1/3 (D) 3/2
Solution: (D) Capacitance (for k 2)
C1 2C Where
Capacitance (for k 6)
C2 3C
Therefore ratio of potential difference across the dielectric layer is
3/2
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