Question:
Two parallel plate capacitors \(8\,\mu\text{F}\) each are connected in parallel to a \(10\,\text{V}\) battery. The plate separation in one of the capacitor is reduced to \(40\%\) of its initial value. The increase in the total charge stored on the capacitors is
Two parallel plate capacitors \(8\,\mu\text{F}\) each are connected in parallel to a \(10\,\text{V}\) battery. The plate separation in one of the capacitor is reduced to \(40\%\) of its initial value. The increase in the total charge stored on the capacitors is
Show Hint
For a parallel plate capacitor,
\[
C\propto \frac{1}{d}
\]
So, reducing the plate separation increases the capacitance proportionally.
Updated On: Jun 26, 2026
- \(80\,\mu\text{C}\)
- \(120\,\mu\text{C}\)
- \(100\,\mu\text{C}\)
- \(150\,\mu\text{C}\)
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The Correct Option is B
Solution and Explanation
Step 1: Find the initial equivalent capacitance.
Two capacitors of capacitance \[ 8\,\mu\text{F} \] each are connected in parallel.
So, the initial equivalent capacitance is \[ C_i=8+8 \] \[ C_i=16\,\mu\text{F} \]
Step 2: Find the initial total charge.
Given battery voltage: \[ V=10\,\text{V} \] Initial total charge stored: \[ Q_i=C_iV \] \[ Q_i=16\times 10 \] \[ Q_i=160\,\mu\text{C} \]
Step 3: Find the new capacitance of the modified capacitor.
For a parallel plate capacitor, \[ C=\frac{\varepsilon_0A}{d} \] Capacitance is inversely proportional to plate separation.
The separation becomes \[ 40\% \] of the original value: \[ d'=0.4d \] Hence new capacitance becomes \[ C'=\frac{C}{0.4} \] \[ C'=2.5C \] Since original capacitance was \[ 8\,\mu\text{F}, \] the modified capacitance is \[ 8\times 2.5 \] \[ =20\,\mu\text{F} \]
Step 4: Find the new equivalent capacitance.
The second capacitor remains unchanged: \[ 8\,\mu\text{F} \] Thus, \[ C_f=20+8 \] \[ C_f=28\,\mu\text{F} \]
Step 5: Find the final total charge.
\[ Q_f=C_fV \] \[ Q_f=28\times 10 \] \[ Q_f=280\,\mu\text{C} \]
Step 6: Calculate the increase in charge.
\[ \Delta Q=Q_f-Q_i \] \[ \Delta Q=280-160 \] \[ \Delta Q=120\,\mu\text{C} \]
Step 7: Final conclusion.
Hence, the increase in total charge stored is \[ \boxed{120\,\mu\text{C}} \]
Two capacitors of capacitance \[ 8\,\mu\text{F} \] each are connected in parallel.
So, the initial equivalent capacitance is \[ C_i=8+8 \] \[ C_i=16\,\mu\text{F} \]
Step 2: Find the initial total charge.
Given battery voltage: \[ V=10\,\text{V} \] Initial total charge stored: \[ Q_i=C_iV \] \[ Q_i=16\times 10 \] \[ Q_i=160\,\mu\text{C} \]
Step 3: Find the new capacitance of the modified capacitor.
For a parallel plate capacitor, \[ C=\frac{\varepsilon_0A}{d} \] Capacitance is inversely proportional to plate separation.
The separation becomes \[ 40\% \] of the original value: \[ d'=0.4d \] Hence new capacitance becomes \[ C'=\frac{C}{0.4} \] \[ C'=2.5C \] Since original capacitance was \[ 8\,\mu\text{F}, \] the modified capacitance is \[ 8\times 2.5 \] \[ =20\,\mu\text{F} \]
Step 4: Find the new equivalent capacitance.
The second capacitor remains unchanged: \[ 8\,\mu\text{F} \] Thus, \[ C_f=20+8 \] \[ C_f=28\,\mu\text{F} \]
Step 5: Find the final total charge.
\[ Q_f=C_fV \] \[ Q_f=28\times 10 \] \[ Q_f=280\,\mu\text{C} \]
Step 6: Calculate the increase in charge.
\[ \Delta Q=Q_f-Q_i \] \[ \Delta Q=280-160 \] \[ \Delta Q=120\,\mu\text{C} \]
Step 7: Final conclusion.
Hence, the increase in total charge stored is \[ \boxed{120\,\mu\text{C}} \]
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