Electrostatics Question 40

Question: Force of attraction between the plates of a parallel plate capacitor is [AFMC 1998]

Options:

A) $ \frac{q^{2}}{2{\varepsilon _{0}}AK} $

B) $ \frac{q^{2}}{{\varepsilon _{0}}AK} $

C) $ \frac{q}{2{\varepsilon _{0}}A} $

D) $ \frac{q^{2}}{2{\varepsilon _{0}}A^{2}K} $

Show Answer

Answer:

Correct Answer: A

Solution:

Force on one plate due to another is F = qE = $ q\times \frac{\sigma }{2{\varepsilon _{0}}K} $

$ =q( \frac{q}{2AK{\varepsilon _{0}}} )=\frac{q^{2}}{2AK{\varepsilon _{0}}} $ (where $ \frac{\sigma }{2{\varepsilon _{0}}K} $ is the electric field produced by one plate at the location of other).



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