1. In the dielectric medium between the plates.

2. High potential, as electrons are negatively charged.

3. Zero

4. Zero

5. ML-1T-2

6. Zero

7. No.

8. Because they are indicators of electric field, extending to infinite distance.

9. C is proportional to A (area) Therefore C2 = 2C1 Since C = Q/V ,So the slope represents more capacitance. Hence P represents C2,Q represents C1

10. Each charge experiences two forces each of magnitude F inclined at an angle of 600.Their

resultant is given by [F2 + F2 +2F2cos 60]1/2 = Ö3.F

11. (i)According to defn of P.D., VP>VQ .So VP-VQ is +ve for q>0.(ii)For q<0, VQ >VP .So VP-VQ is -ve

12. Flux = 0, since Qen = 0

13. Due to polaraisation,opposing electric field is created.

14. Electric field at the midpoint of a dipole of length 2a is 2kq/a pointing towards the –ve charge or in the direction opposite to the dipole moment.

15. Inside the cavity field at any point is uniform and non zero.

16. No. If the initial velocity of the charged particle makes a certain angle with a line of force, then the charged particle shall not move along the line of force.

17. E= -dv/dr= -d(q/4πεor)/dr = q/4πεor2

18. Yes, at the mid point of electric dipole.

19. U = kq2/a- kq2/2a- kq2/2a = 0.

20. C’ = C+C+C = 3C = 75μF Therefore, charge = 75μF x 4200 V = 315 mC*

21. Total ф = q0/εo = 2/εo ,→ flux through one face = ф/6 = 1/3 εo.

22. q →------------ Q

1/2mv2 = kQq/r

Or, v2 α 1/r

Or, r α 1/v2

Or, r’ = r/4

23. V = V1+V2

Q =C1V = 6 X 10-6 X 2 = 12mC

As C2 is in series same amount of charge will also flow through it. Now V2= Q/ C2 = (12 X 10-6)/ (12 X 10-6) = 1 Volt Total battery Voltage,V = 2 + 1 = 3 Volt

24.Capacitance of parallel plate capacitor with air between the plates is C0 = e0A/d When the separation between the plates reduced to half, C1=e0A/(d/2) = 2e0A/d Thus final capacitance is C2=10 X 8 pF=80 pF

25.The arrangement is of 5 capacitors in series.Therefore 1/C’= (1/C) +(1/C) +(1/C) +(1/C) +(1/C) =(5/C) Therefore C’=C/5 Or 5=C/5 or C=25mF.

26. The charge given to a capacitor is given by q=CV So the remaining energy,qV- 1/2 qV =1/2 qV is lost as heat

27. On equatorial line ,the direction of electric field is reversed to that of axial line. Hence the angle between electric dipole moment& electric field strength is 1800

CBSE Physics (Chapter Wise With Hint / Solution) Class XII (By Mr. Sreekumaran Nair)