JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 ...... \(N\)
- A \(5\)
- B \(10\)
- C \(20\)
- D \(0\)
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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