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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2022
  • Medium
  • Numerical

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 ...... \(N\)

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Answer & Solution

Correct Answer

(A) \(5\)

Step-by-step Solution

Detailed explanation

\(F=q E=q\left(\frac{Q}{A \epsilon_{0}}\right)=\frac{q Q}{A \epsilon_{0}}=10 \; N\)

Now, when one plate is removed.

\(E^{\prime}=\frac{Q}{2 A \epsilon_{0}}\)

\(F=q E^{\prime}=\frac{Q q}{2 A \epsilon_{0}}=5 \; N\)

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