SATHEE: Electrostatics Question 264

Electrostatics Question 264

Question: A dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 0$ and positive plate is at $x = 3 d$ . The slab is equidistant from the plates. The capacitor is given some charge. As one goes from 0 to $3 d$ [IIT-JEE 1998]

Options:

A) r

B) 2r

C) r/2

D) r/4

Show Answer

Answer:

Correct Answer: D

Solution:

Charge q will momentarily come to rest at a distance r from charge Q when all it’s kinetic energy converted to potential energy i.e. $\frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} . \frac{q Q}{r}$

Therefore the distance of closest approach is given by $r = \frac{q Q}{4 π ε_{0}} . \frac{2}{m v^{2}}$

therefore $r \propto \frac{1}{v^{2}}$

Hence if v is doubled, r becomes one fourth.

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Electrostatics Question 264

Question: A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x=0 and positive plate is at x=3d . The slab is equidistant from the plates. The capacitor is given some charge. As one goes from 0 to 3d [IIT-JEE 1998]

Options:

Answer:

Solution:

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Question: A dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 0$ and positive plate is at $x = 3 d$ . The slab is equidistant from the plates. The capacitor is given some charge. As one goes from 0 to $3 d$ [IIT-JEE 1998]