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:
r
2r
r/2
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}}}\cdot\frac{1}{mv^{2}} $
therefore $ r\propto \frac{1}{v^{2}} $
Hence if v is doubled, r becomes one fourth.
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