Question

Question: (II) An uncharged capacitor is connected to a 21.0-V battery until it is fully charged, after which it is disconnected from the battery. A slab of paraffin is then inserted between the plates. What will now be the voltage between the plates?

Short answer

The voltage between the plates is 9.5 V.

Step-by-step solution

Understanding the effect of dielectric on capacitance

The capacitance of a capacitor relies on the area of capacitor plates and separation between the plates. The value of capacitor increases when a dielectric material is inserted between the plates.

The expression for the capacitor is given as:

$C={\epsilon }_0\frac{A}{d}$ … (i)

Here,${\epsilon }_0$is the permittivity of free space, A is the area of plate and d is the separation between plates.

The charge on the capacitor remains the same after disconnecting the battery.

Given data

The voltage of the battery is,$V=21.0\mathrm{V}$

The dielectric constant of paraffin is $K=2.2$.

Determination of the final voltage across the plates

The charge on the capacitor is,

$Q=CV$

After disconnecting from the battery, the charge on the plates remains same. Initially, there is air between the plates.

The capacitance of the capacitor when paraffin is inserted between the plates is,

$\begin{array}{rl}l{C}^{\prime }& =K{\epsilon }_0\frac{A}{d}\\ C^{\prime }& =KC\end{array}$

As the charge remains the same then,

$\begin{array}{rl}c{C}^{\prime }{V}^{\prime }& =Q\\ C^{\prime }{V}^{\prime }& =CV\\ KC{V}^{\prime }& =CV\\ V^{\prime }& =\frac{V}{K}\end{array}$

Substitute the values in the above expression.

$\begin{array}{rl}c{V}^{\prime }& =\frac{21.0\mathrm{V}}{2.2}\\ & =9.54\mathrm{V}\\ \approx 9.5\mathrm{V}\end{array}$

Thus, the voltage between the plates is 9.5 V.