A parallel plate capacitor is formed by two plates each of area 30π cm2 separated by 1 mm. A material of dielectric strength 3.6 × 107 V/m is filled between the plates. If the maximum charge that can be stored in the capacitor without causing any dielectric breakdown is 7 × 10-6 C, the value of dielectric constant of the material is
(Given: \( \displaystyle\frac{1}{4\pi\varepsilon_0} \) = 9 × 109 Nm2/C2 )
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2.33
The dielectric strength is the largest electric field the material can withstand without breaking down. With a dielectric of constant K between the plates, the field is \( E = \sigma/K\varepsilon_0 = Q/K\varepsilon_0 A \), so the maximum charge is reached when this field equals the dielectric strength.
\( \displaystyle E_{\mathrm{max}} = \frac{Q_{\mathrm{max}}}{K\varepsilon_0 A} \)
\( \displaystyle K = \frac{Q_{\mathrm{max}}}{\varepsilon_0 A E_{\mathrm{max}}} \)
Convert the area and write \( \varepsilon_0 \) from the given constant:
\( A = 30\pi\,\mathrm{cm^{2}} = 30\pi\times10^{-4}\,\mathrm{m^{2}} \)
\( \displaystyle \varepsilon_0 = \frac{1}{4\pi \times 9\times10^{9}}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}} \)
\( \displaystyle \varepsilon_0 A = \frac{30\pi\times10^{-4}}{36\pi\times10^{9}} = \frac{5}{6}\times10^{-13}\,\mathrm{C^{2}\,N^{-1}} \)
The factor \( \pi \) cancels, which is why the area was given as 30π cm2.
\( \displaystyle \varepsilon_0 A E_{\mathrm{max}} = \frac{5}{6}\times10^{-13} \times 3.6\times10^{7} = 3.0\times10^{-6}\,\mathrm{C} \)
\( \displaystyle K = \frac{7\times10^{-6}\,\mathrm{C}}{3.0\times10^{-6}\,\mathrm{C}} \approx 2.33 \)
The plate separation (1 mm) is not needed, because the breakdown limit is set by the field, not by the voltage.