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

• A. \(80\,\mu\text{C}\)
• B. \(120\,\mu\text{C}\)
• C. \(100\,\mu\text{C}\)
• D. \(150\,\mu\text{C}\)

Hint:

For a parallel plate capacitor, \[ C\propto \frac{1}{d} \] So, reducing the plate separation increases the capacitance proportionally.

Answer 1

The Correct Option is B

Step 1: Find the initial total capacitance and charge.
Two capacitors of $8\,\mu\text{F}$ each are connected in parallel to a $10\,\text{V}$ battery. Capacitors in parallel add directly: \[ C_i = 8 + 8 = 16\,\mu\text{F} \] Initial total charge: \[ Q_i = C_i V = 16 \times 10 = 160\,\mu\text{C} \]
Step 2: Understand the effect of reducing plate separation.
For a parallel plate capacitor, $C = \frac{\varepsilon_0 A}{d}$. When the plate separation $d$ decreases, the capacitance increases. If separation becomes $40\%$ of original (i.e., $d' = 0.4d$), the new capacitance is: \[ C' = \frac{\varepsilon_0 A}{0.4d} = \frac{C}{0.4} = 2.5C \]
Step 3: Calculate the new capacitance of the modified capacitor.
The original capacitance of the modified capacitor was $8\,\mu\text{F}$. After reducing plate separation to $40\%$: \[ C'_1 = 2.5 \times 8 = 20\,\mu\text{F} \] The other capacitor remains unchanged at $8\,\mu\text{F}$.
Step 4: Find the new total capacitance and charge.
Both are still connected in parallel to the same $10\,\text{V}$ battery, so: \[ C_f = 20 + 8 = 28\,\mu\text{F} \] \[ Q_f = C_f V = 28 \times 10 = 280\,\mu\text{C} \]
Step 5: Calculate the increase in charge.
\[ \Delta Q = Q_f - Q_i = 280 - 160 = 120\,\mu\text{C} \] Since the battery is still connected, it supplies additional charge to maintain $10\,\text{V}$ as the capacitance increases.
Step 6: State the final answer.
The increase in total charge stored is: \[ \boxed{120\,\mu\text{C}} \]