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.
Solution
<p>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</p>
<p>So new force F = 5 N</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Field and Field Lines
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