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 \(\underline{\hspace{2cm}}\).

Hint:

When a capacitor is disconnected before removing dielectric, charge stays constant but electric field scales with permittivity.

Answer 1

Correct Answer: 5

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:
\[ \frac{E_2}{E_1} = \varepsilon_r = 5 \] \[ \boxed{5} \]