Question

A parallel plate capacitor of capacitance \(5\ \mu F\) and plate separation \(6\ \text{cm}\) is connected to a \(1\ \text{V}\) battery and charged. A dielectric of dielectric constant \(4\) and thickness \(4\ \text{cm}\) is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is:

• A. \(2\ \mu C\)
• B. \(3\ \mu C\)
• C. \(5\ \mu C\)
• D. \(10\ \mu C\)

Hint:

Remember: \[ d_{\text{eq}}=(d-t)+\frac{t}{K} \]

• Dielectric insertion increases capacitance
• If battery remains connected: \[ Q=CV \] changes because \(C\) changes

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When a capacitor remains continuously connected to a battery source, its potential difference ($V$) stays fixed at the battery voltage. Introducing a non-conducting dielectric material partially into the gap increases the absolute capacitance ($C'$). To maintain the constant voltage balance at this higher capacitance, the battery must supply an additional quantity of charge ($\Delta Q$) to the plates.
Step 2: Key Formula or Approach:
- Initial Charge: $Q_i = C_i V$ - Capacitance with a partially filled dielectric slab of thickness $t$: \[ C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}} \] - Additional charge flowing from the battery: $\Delta Q = Q_f - Q_i = (C' - C_i)V$
Step 3: Detailed Explanation:
Given values: - Initial capacitance, $C_i = 5 \ \mu\text{F}$ - Total plate separation, $d = 6 \text{ cm}$ - Dielectric thickness, $t = 4 \text{ cm}$ - Dielectric constant, $K = 4$ - Battery voltage, $V = 1 \text{ V}$ Let's express the initial capacitance in terms of area and distance: \[ C_i = \frac{\varepsilon_0 A}{d} = 5 \ \mu\text{F} \implies \varepsilon_0 A = 5 \times d \] Now write out the formula for the modified new capacitance $C'$: \[ C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}} \] Substitute the given values for $d$, $t$, and $K$: \[ C' = \frac{\varepsilon_0 A}{6 - 4 + \frac{4}{4}} = \frac{\varepsilon_0 A}{2 + 1} = \frac{\varepsilon_0 A}{3} \] We know from our initial condition that $\varepsilon_0 A = C_i \times d = 5 \times 6 = 30 \ \mu\text{F} \cdot \text{cm}$. Substitute this back to evaluate $C'$: \[ C' = \frac{30}{3} = 10 \ \mu\text{F} \] Now calculate the change in charge ($\Delta Q$): - Initial Charge: $Q_i = C_i V = 5 \ \mu\text{F} \times 1 \text{ V} = 5 \ \mu\text{C}$ - Final Charge: $Q_f = C' V = 10 \ \mu\text{F} \times 1 \text{ V} = 10 \ \mu\text{C}$ - Additional Charge: $\Delta Q = Q_f - Q_i = 10 \ \mu\text{C} - 5 \ \mu\text{C} = 5 \ \mu\text{C}$
Step 4: Final Answer:
The additional charge that flows into the capacitor from the battery is 5 µC.