Electrostatics Question 49
Question: Force acting upon a charged particle kept between the plates of a charged condenser is $ F $ . If one plate of the condenser is removed, then the force acting on the same particle will become [MP PMT 1991]
Options:
A) 0
B) . $ F/2 $
C) $ F $
D) $ 2F $
Show Answer
Answer:
Correct Answer: B
Solution:
Initially F = qE and $ E=\frac{\sigma }{{\varepsilon _{0}}} $
$ F=\frac{q\sigma }{{\varepsilon _{0}}} $
If one plate is removed, then E becomes $ \frac{\sigma }{2{\varepsilon _{0}}} $ So $ F’=\frac{q\sigma }{2{\varepsilon _{0}}}=\frac{F}{2} $
