Question:
A parallel-plate capacitor with dielectric of relative permittivity 5 is charged to \(V\) and disconnected. The slab is pulled out completely. The ratio of new electric field \(E_2\) to original field \(E_1\) is \(\underline{\hspace{2cm}}\).
A parallel-plate capacitor with dielectric of relative permittivity 5 is charged to \(V\) and disconnected. The slab is pulled out completely. The ratio of new electric field \(E_2\) to original field \(E_1\) is \(\underline{\hspace{2cm}}\).
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When a capacitor is disconnected before removing dielectric, charge stays constant but electric field scales with permittivity.
Updated On: Dec 29, 2025
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Correct Answer: 5
Solution and Explanation
Initially with dielectric:
\[ E_1 = \frac{V}{d} \cdot \frac{1}{\varepsilon_r} \] After removing dielectric (air gap):
\[ E_2 = \frac{V}{d} \] Thus ratio:
\[ \frac{E_2}{E_1} = \varepsilon_r = 5 \] \[ \boxed{5} \]
\[ E_1 = \frac{V}{d} \cdot \frac{1}{\varepsilon_r} \] After removing dielectric (air gap):
\[ E_2 = \frac{V}{d} \] Thus ratio:
\[ \frac{E_2}{E_1} = \varepsilon_r = 5 \] \[ \boxed{5} \]
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