Question

A parallel plate capacitor is formed by two plates each of area \(30\pi\,\text{cm}^2\) separated by \(1\,\text{mm}\). A material of dielectric strength \(3.6\times10^7\,\text{V m}^{-1}\) is filled between the plates. If the maximum charge that can be stored in the capacitor without causing any dielectric breakdown is \(7\times10^{-6}\,\text{C}\), the value of dielectric constant of the material is

• A. \(1.66\)
• B. \(1.75\)
• C. \(2.25\)
• D. \(2.33\)

Hint:

At dielectric breakdown: \[ Q_{\max}=K\varepsilon_0AE_{\max} \] Notice that the plate separation \(d\) cancels out. This shortcut is very useful in capacitor MCQs involving dielectric strength.

Answer 1

The Correct Option is D

Concept: For a parallel plate capacitor, \[ Q=CV \] where \[ C=\frac{K\varepsilon_0A}{d} \] At dielectric breakdown, \[ V_{\max}=E_{\max}d \] Therefore, \[ Q_{\max} = \frac{K\varepsilon_0A}{d} (E_{\max}d) = K\varepsilon_0AE_{\max} \]

Step 1: Write the given data. \[ Q_{\max}=7\times10^{-6}\,\text{C} \] \[ E_{\max}=3.6\times10^7\,\text{V m}^{-1} \] \[ A=30\pi\,\text{cm}^2 \] \[ A=30\pi\times10^{-4} =3\pi\times10^{-3}\,\text{m}^2 \] Also, \[ \frac{1}{4\pi\varepsilon_0}=9\times10^9 \] Hence, \[ \varepsilon_0=\frac{1}{36\pi\times10^9} \]

Step 2: Substitute into the formula. \[ Q_{\max} = K\varepsilon_0AE_{\max} \] \[ 7\times10^{-6} = K \left(\frac{1}{36\pi\times10^9}\right) \left(3\pi\times10^{-3}\right) \left(3.6\times10^7\right) \] \[ 7\times10^{-6} = K\times3\times10^{-6} \] \[ K=\frac{7}{3} \] \[ K=2.33 \]

Step 3: State the answer. \[ \boxed{ K=2.33 } \] Hence, the correct option is \[ \boxed{(D)} \]