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Q. No.An electron (charge: \(e\) & mass: \(m\)) emitted from the positive plate \(A\) of the capacitor \(AB\), just manages to reach the negative plate \(B\). If the potential difference between the plates is \(V_0\), then the speed of the electron when it is emitted is:
| 1. | \(\sqrt{\frac{e V_{0}}{m}} \quad\) | 2. | \(\sqrt{\frac{e V_{0}}{2 m}} \quad\) |
| 3. | \(\sqrt{\frac{2 e V_{0}}{m}}\) | 4. | \(\sqrt{\frac{2 e V_{0} d}{m}}\) |
Subtopic: Â Electric Potential Energy |
From NCERT
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