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 objective is to determine the supply voltage applied to a parallel plate capacitor that has been modified with dielectric materials. The initial capacitance and dimensions, the dielectric constants of the new materials, and the resultant attractive force between the plates are provided.

Concept Used:

The solution employs several fundamental concepts from electrostatics:

1. Capacitors in Parallel: When the inter-plate space of a capacitor is filled with two dielectric slabs of equal area side-by-side, this configuration is equivalent to two capacitors connected in parallel.
2. Effect of Dielectrics on Capacitance: The insertion of a dielectric material with constant \(K\) into a capacitor results in its original capacitance being multiplied by \(K\). For a capacitor with area \(A\) and separation \(d\), the capacitance of a portion with dielectric \(K_1\) and area \(A/2\) is \(C_1 = \frac{K_1 \epsilon_0 (A/2)}{d}\).
3. Equivalent Capacitance: The total capacitance of capacitors connected in parallel is the sum of their 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 defined by the stored energy \(U\) and the plate separation \(d\) as \(F = \frac{U}{d}\). For a capacitor connected to a voltage source \(V\), the stored energy is \(U = \frac{1}{2} C_{eq} V^2\). Combining these yields the force formula: \[ F = \frac{1}{2d} C_{eq} V^2 \]

Step-by-Step Solution:

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

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

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

Step 2: Compute 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 segments are:

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

Since these two segments are in parallel, the new equivalent capacitance is their sum:

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

Step 3: Substitute the given values to determine 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: Employ the force formula to calculate the supply voltage \(V\).

The relationship between force, capacitance, voltage, and plate separation is given by:

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

Rearrange this equation to solve for voltage \(V\):

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

Step 5: Substitute the known values and compute 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 required supply voltage is 80 V.