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:

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}mv^{2}=\frac{1}{4\pi {\varepsilon _{0}}}.\frac{qQ}{r} $

Therefore the distance of closest approach is given by $ r=\frac{qQ}{4\pi {\varepsilon _{0}}}.\frac{2}{mv^{2}} $

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

Hence if v is doubled, r becomes one fourth.



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