Question

A capacitor of $10 \, \mu\text{F}$ capacitance whose plates are separated by $10 \, \text{mm}$ through air and each plate has area $4 \, \text{cm}^2$ is now filled equally with two dielectric media of $K_1 = 2$, $K_2 = 3$ respectively as shown in the figure. If new force between the plates is $8 \, \text{N}$, the supply voltage is ____ $\text{V}$.

Answer 1

Correct Answer: 80

The problem asks for the supply voltage applied to a parallel plate capacitor that has been modified with dielectric materials. We are given the initial capacitance and dimensions, the dielectric constants of the new materials, and the resulting attractive force between the plates.

Concept Used:

The solution involves several key concepts from electrostatics:

1. Capacitors in Parallel: When the space between the plates of a capacitor is filled with two dielectric slabs side-by-side, each occupying half the area, the arrangement is equivalent to two separate capacitors connected in parallel.
2. Effect of Dielectrics on Capacitance: Inserting a dielectric of constant \(K\) into a capacitor multiplies its original capacitance by \(K\). For a capacitor with area \(A\) and separation \(d\), the capacitance of one part with dielectric \(K_1\) and area \(A/2\) is \(C_1 = \frac{K_1 \epsilon_0 (A/2)}{d}\).
3. Equivalent Capacitance: The equivalent capacitance of a parallel combination is the sum of the individual capacitances: \(C_{eq} = C_1 + C_2\).
4. Force Between Capacitor Plates: The magnitude of the attractive force \(F\) between the plates of a parallel plate capacitor is related to the energy stored \(U\) and the plate separation \(d\) by the formula \(F = \frac{U}{d}\). When the capacitor is connected to a voltage source \(V\), the stored energy is \(U = \frac{1}{2} C_{eq} V^2\). Combining these gives the force as: \[ F = \frac{1}{2d} C_{eq} V^2 \]

Step-by-Step Solution:

Step 1: List the given parameters and model the new capacitor configuration.

• Initial capacitance (with air), \(C_{\text{air}} = 10 \, \mu\text{F} = 10 \times 10^{-6} \, \text{F}\).
• Plate separation, \(d = 10 \, \text{mm} = 10 \times 10^{-3} \, \text{m}\).
• Dielectric constants, \(K_1 = 2\) and \(K_2 = 3\).
• New force between the plates, \(F = 8 \, \text{N}\).

The modified capacitor can be treated as two capacitors in parallel. Let \(C_1\) be the capacitor with dielectric \(K_1\) and area \(A/2\), and \(C_2\) be the one with \(K_2\) and area \(A/2\).

Step 2: Calculate the new equivalent capacitance \(C_{\text{new}}\).

The capacitance of the original air-filled capacitor is \(C_{\text{air}} = \frac{\epsilon_0 A}{d}\). The capacitances of the two new sections are:

\[ C_1 = \frac{K_1 \epsilon_0 (A/2)}{d} = \frac{K_1}{2} \left(\frac{\epsilon_0 A}{d}\right) = \frac{K_1}{2} C_{\text{air}} \] \[ C_2 = \frac{K_2 \epsilon_0 (A/2)}{d} = \frac{K_2}{2} \left(\frac{\epsilon_0 A}{d}\right) = \frac{K_2}{2} C_{\text{air}} \]

The new equivalent capacitance is the sum of these two, as they are in parallel:

\[ C_{\text{new}} = C_1 + C_2 = \frac{K_1}{2} C_{\text{air}} + \frac{K_2}{2} C_{\text{air}} = \frac{C_{\text{air}}}{2} (K_1 + K_2) \]

Step 3: Substitute the given values to find the numerical value of \(C_{\text{new}}\).

\[ C_{\text{new}} = \frac{10 \times 10^{-6} \, \text{F}}{2} (2 + 3) \] \[ C_{\text{new}} = (5 \times 10^{-6}) \times 5 = 25 \times 10^{-6} \, \text{F} = 25 \, \mu\text{F} \]

Step 4: Use the force formula to find the supply voltage \(V\).

The formula relating force, capacitance, voltage, and plate separation is:

\[ F = \frac{1}{2d} C_{\text{new}} V^2 \]

We can rearrange this equation to solve for the voltage \(V\):

\[ V^2 = \frac{2Fd}{C_{\text{new}}} \] \[ V = \sqrt{\frac{2Fd}{C_{\text{new}}}} \]

Step 5: Substitute the known values and calculate the final voltage.

\[ V = \sqrt{\frac{2 \times 8 \, \text{N} \times (10 \times 10^{-3} \, \text{m})}{25 \times 10^{-6} \, \text{F}}} \] \[ V = \sqrt{\frac{160 \times 10^{-3}}{25 \times 10^{-6}}} = \sqrt{\frac{160}{25} \times 10^3} \] \[ V = \sqrt{6.4 \times 1000} = \sqrt{6400} \] \[ V = 80 \, \text{V} \]

The supply voltage is 80 V.