Electrostat-09-Solution Subjective and Objective

ELECTROSTATIC (SOLUTIONS) . Let potential of ‘O’ is Then (Using Kirchhoff’s Law) Solving 2. loss in P.E. = Gain in K.E. = = 3. Considering a small segment ‘dx’ at distance ‘x’ from hinged ‘A’ charge appearing on it. force Torque Total torque …(i) Now (Using ) unit 4. Potential due to one drop Radius ‘R’ of bigger drop = R = 10 r …(i) New potential 5. After touching one of the charged ball charge on uncharged ball will half of it. Let charge on each ball be q. …(i) After touching 6. As per the question …(i) when outer sphere is earthed, charge appearing on it is –Q Now potential of point P 7. Let ‘n’ such capacitors are in series and such ‘m’ such branch are in parallel. 250×n = 1000 n = 4 …(i) Also = 8 …(ii) No. of capacitor = 8. Charge on capacitor ‘c’ = CV While on 2 C = 2CV Total charge = 3CV …(i) Since capacitor ‘C’ is filled with dielectric of constant ‘K’. New capacitance is CK Let and be final charge on ‘C’ and ‘2C’ …(ii) but solving New p.d. = 9. Now Force Nt 10. Potential at sight Similarly 11. ( are in series and resultant of these two in parallel with C3) 12. After disconnecting battery charges on capacitors is CV and 2CV respectively. Let final charges on capacitors C and 2c is ‘q1’ and ‘q2’ (using charge conservation) New P.D = 13. Let x charge is given to inner shell 14. (which is sinusoidal). for , (max.) 15. Charge on sphere A …(i) After connecting with sphere ‘B’ net charge on ‘A’ will be zero and total charge will reside on B only Potential at sight of A 1. …(i) …(ii) loss in energy = 2. Initial potential energy of the system = 0 …(i) The rod will rotate in clockwise sense. When rod becomes horizontal …(ii) loss in P.E. = (using ) 3. Initial potential energy …(i) …(ii) Work done by external agent = = 4. Initial charge on inner shell is zero. When it is earthed its potential must be zero. Let final charge on it is x Then (final charge) charge flown = 5. Initial charge on capacitor and final charge i.e. there is no charge in stored energy of capacitor. Total charge supplied by the battery in second case = 2CE Work done by battery = ( ) 6. Let radius of inner, middle and outer sphere be and respectively and final charge on outer sphere be y then final charge distribution is given below i.e. outer surface of outer sphere will remains chargeless. (a) is correct 7. We have two capacitors and After connecting then with a conducting wire they will be in parallel grouping. 8. Energy stored in capacitor where is constant power supply where V is speed of plate which is constant 9. Initial charge Let after connective ‘C’ charge appearing on it is Then charge left on Solving Charge left on For n such connection charge left on = P.D. (as per equation) Solving we get 10. 11. Let final charge on 1 F capacitor is x. Using charge conservation, charge distribution is shown in above diagram Using Kirchhoff’s law (charge on 2 ) and (charge on 3 ) Ans. (c) 12. We have two capacitor (between A and B) and (between C and D) = and they are In parallel 13. The charge on plate x is bounded due to negative charge on the rod. This particular charge will be +ve in nature and it can’t flow to earth. 14. Surface BPD is equipotential Here electric field must be to BPD and along CA. (E is directed along decreasing potential) length cm Here V/cm. 15. also 1. (a) Putting the respective values (b) Using Newton’s third law force will be equal and opposite. 2. (at P(1,1,1) 3. (a) along OB (b) Force on –ve charge 1.5×10–9C is along BA (Using F = qE0) = 81× 10-5 N along BA 4. Total charge = 0 and dipole moment = cm 5. 1st and 2nd charges are –ve because they are deviated towards +ve plate and 3rd charge is +vely charged because it is deviated towards –ve plate. Charge/mass is greatest for 3rd because it is suffering maximum deviation. 6. (a) Flux (b) 7. Let us consider a cube of side 10 cm which encloses charge +10 mc Total flux (six surfaces) = Flux through one of side = 8.(a) Equipotential surface is a surface where potential has constant valve. Consider a point P where potential is zero. SUBJECTVE LEVEL I r1 = r2 A plane perpendicular to AB can be equipotential (b) Resultant field will be to equipotential surface along AB. 9. (a) for a point just inside the sphere no charge is enclosed (b) For a point just outside (R radius of sphere) = (c) At a point 18 cm 10. (a) Equivalent capacitance (b) 11. (a) (b) From summitry field at center will be zero. 12. 13. (a) (b) 14. (a) = 9 pF (b) 15 work done . SUBJECTIVE LEVEL II …(i) New time period Total time taken to have 20 oscillations 2. Field due to ring at distance ‘x’ from its center and on its axis. where R is radius. For x << R force attractive Here (restoring in nature) 3. Let mass of the bob be m …(i) Force of buoyancy 1st condition …(ii) 2nd condition In equilibrium …(iii) (where K is dielectric constant of medium) dividing eq. (ii) and (iii) 4. Let charge on second particle ‘B’ is …(i) maximum friction that can act on the particle …(ii) For equilibrium to exist . 5. Equivalent capacitance eq. = Total charge supplied by the battery …(i) charge on 8 capacitor = 48 while equal charge will be distributed over each of 4 capacitors i.e. charge on each 4 capacitors 6. Ceq (when switch is open) = when switch is closed, capacitor across it will be short circuited Additional charge flown 250–125 = 125 mC 7. Electric field The charge will remaining equilibrium If putting respective values 8. …(i) After inserting dielectric with open switch incharged condition …(ii) 9. Consider a thin strip at a distance x and of thickness dx Area of strip = adx Distance between strip = d + x 10. Let at any time separation between two upper plate is x. then separation between two tower plate will be = which is independent of position of central section. SUBJECTIVE LEVEL III 1. Here there will not be any flux through the surfaces whose area vector is perpendicular to x-axis Flux through surface ADSP while flux through BCRQ (b) 2.(a) , (b) Here with (c) 3. Considering a ring of radius x and thickness ‘dx’ putting Total flux 4.(a) Charge in upper branch = = Now Similarly (b) Since No charge will accumulate on this capacitor 5. Here we have two capacitor (a) when out sphere ‘B is earthed (b) When inner sphere is earthed both capacitors are in parallel F 6.(a) If charge q remains constant work (b) when P.D is constant force is not constant 7. Here …(i) when there is no dielectric as per the question i.e. solving 8.(a) Using charge distribution At point P, electric filled must be zero i.e. charge on outer surface of upper and lower plate is (b) P.d. = 9. We have two capacitor is in parallel If da is further increment in length of capacitor inside capacitor let de is increment in capacitance Using conservation of energy (independent of ‘a’) time taken to cover Total time for one complete oscillation 10.(a) In equilibrium spring force balanced by force between plates. i.e. (b) Let at any time t separation between plates is ‘x’ then …(i) differentiating w.r.t. x …(ii) Putting the value of ‘x’ from equation (ii) in equation (i)