JEE MAIN - Physics (2022 - 27th June Morning Shift - No. 14)
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.
5 N
10 N
20 N
Zero
Spiegazione
E between two plates is $${\sigma \over {{\varepsilon _0}}}$$ and due to one plate is $${\sigma \over {2{\varepsilon _0}}}$$ so the force will be halved
So new force F = 5 N