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).
BETA
