Question

A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be

• A. 5 N
• B. 10 N
• C. 20 N
• D. Zero

Answer 1

Answer: (A)

5 N

Answer 2

E between two plates is $\frac{σ}{ε_0}$

and due to one plate is $\frac{σ}{2ε_0}$

Therefore, the force will be half of the original force

New force $F = 5$ $N$.

Hence, the correct option is (A) : $5 \;N$