Consider a large parallel plate capacitor. The gap d between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of V volts and then disconnected from the source. If the dielectric slab is pulled out completely, then the ratio of the new electric field E₂ in the gap to the original electric field E₁ is __________.

Correct Answer: The correct answer is 5.

Step-by-Step Explanation:

1. Initial State (Dielectric Present):

Let the parallel plate capacitor have a plate area $A$ and separation gap $d$.

Initially, the gap is filled completely with a dielectric slab of relative permittivity (dielectric constant) ${\epsilon }_r=5$.

The initial capacitance $C_1$ is given by:

$C_1=\frac{5{\epsilon }_{0}A}{d}$

The plates are charged to a potential difference of $V$ volts. The charge stored on the plates initially is:

$Q=C_{1}V$

The original electric field inside the dielectric slab, $E_1$, is related to the potential difference $V$ and distance $d$ by:

$E_1=\frac{V}{d}=\frac{\sigma }{5{\epsilon }_0}$

where $\sigma =\frac{Q}{A}$ is the surface charge density on the plates.

2. Final State (Dielectric Removed):

The charging source is disconnected after initial charging. Therefore, the charge $Q$ (and consequently the surface charge density $\sigma$) remains constant on the plates because there is no path for the charge to escape.

When the dielectric slab is completely pulled out, the space between the plates becomes air/vacuum with relative permittivity ${\epsilon }_r=1$.

The new electric field in the gap, $E_2$, depends only on the free surface charge density $\sigma$ on the conductive plates:

$E_2=\frac{\sigma }{{\epsilon }_0}$

3. Ratio of Electric Fields:

Taking the ratio of the new electric field $E_2$ to the original electric field $E_1$:

$\frac{E_2}{E_1}=\frac{\sigma /{\epsilon }_0}{\sigma /\left(5{\epsilon }_0\right)}=5$

Thus, the ratio of the new electric field $E_2$ to the original electric field $E_1$ is 5.