Question:
In a parallel plate capacitor, if \(10^{12}\) electrons pass from one plate to another, a potential difference of \(10\,\text{V}\) is developed across the plates. The capacitance of the capacitor is:
In a parallel plate capacitor, if \(10^{12}\) electrons pass from one plate to another, a potential difference of \(10\,\text{V}\) is developed across the plates. The capacitance of the capacitor is:
Show Hint
For capacitor problems involving transferred electrons, first calculate charge using
\[
Q=ne
\]
and then use
\[
C=\frac{Q}{V}.
\]
Updated On: Jun 26, 2026
- \(0.16\times 10^{-8}\,\text{F}\)
- \(1.6\times 10^{-8}\,\text{F}\)
- \(16\times 10^{-8}\,\text{F}\)
- \(0.8\times 10^{-8}\,\text{F}\)
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The Correct Option is B
Solution and Explanation
Step 1: Calculate the charge transferred.
When \(10^{12}\) electrons are transferred from one plate to another, the magnitude of charge on each plate becomes \[ Q=ne \] where \[ n=10^{12} \] and \[ e=1.6\times 10^{-19}\,\text{C} \] Therefore, \[ Q=(10^{12})(1.6\times 10^{-19}) \] \[ Q=1.6\times 10^{-7}\,\text{C} \]
Step 2: Use the capacitance formula.
Capacitance is defined as \[ C=\frac{Q}{V} \] Given, \[ V=10\,\text{V} \] Substituting the values, \[ C=\frac{1.6\times 10^{-7}}{10} \] \[ C=1.6\times 10^{-8}\,\text{F} \]
Step 3: Verify with the given options.
The obtained value is \[ 1.6\times 10^{-8}\,\text{F} \] which matches option (2).
Step 4: Final conclusion.
Therefore, the capacitance of the capacitor is \[ \boxed{1.6\times 10^{-8}\,\text{F}} \]
When \(10^{12}\) electrons are transferred from one plate to another, the magnitude of charge on each plate becomes \[ Q=ne \] where \[ n=10^{12} \] and \[ e=1.6\times 10^{-19}\,\text{C} \] Therefore, \[ Q=(10^{12})(1.6\times 10^{-19}) \] \[ Q=1.6\times 10^{-7}\,\text{C} \]
Step 2: Use the capacitance formula.
Capacitance is defined as \[ C=\frac{Q}{V} \] Given, \[ V=10\,\text{V} \] Substituting the values, \[ C=\frac{1.6\times 10^{-7}}{10} \] \[ C=1.6\times 10^{-8}\,\text{F} \]
Step 3: Verify with the given options.
The obtained value is \[ 1.6\times 10^{-8}\,\text{F} \] which matches option (2).
Step 4: Final conclusion.
Therefore, the capacitance of the capacitor is \[ \boxed{1.6\times 10^{-8}\,\text{F}} \]
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