**Calculate the potential difference existing between the plates of the capacitor.****Keeping the charge constant on each plate of the capacitor, calculate the impact of doubling the separation between capacitor plates on the potential difference.****Calculate the amount of work that will be required to double the separation between the capacitor plates.**

The aim of this article is to find the **potential difference** between the **capacitor plates** having a certain **charge** and the impact of changing the **separation** between the **capacitor plates** on the **potential difference** and the **work done** to execute it.

The main concept behind this article is the understanding of **Charge on Capacitor** Q, **Capacitance of the Capacitor** C, and the **Work done** W in relation to **Potential difference**V across the** capacitor plates**.

**Charge on Capacitor** $Q$, **Capacitance of the Capacitor** $C$ and the **Work done** $W$ in relation to **Potential difference** $V$ across the **capacitor plates** are expressed as the following relation:

**Charge on the Capacitor** $Q$ is:

\[Q=CV\]

Where:

$Q=$ **Charge on Capacitor Plates**

$C=$ **Capacitance of Capacitor**

$V=$ **Potential Difference across capacitor plates**

The **Capacitance of the Capacitor** $C$ is:

\[C=\frac{\varepsilon_oA}{d}\]

Where:

$C=$ **Capacitance of Capacitor**

$\varepsilon_o=$ **Permittivity of Free Space**

$A=$ **Area of the Parallel Plates of**

$d=$ **Separation between the Capacitor Plates**

**Work Done** to increase the **separation** between the **capacitor plates** $W$ is:

\[W=\frac{1}{2}QV\]

## Expert Answer

Given that:

**Capacitance of Capacitor** $C=920pF=920\times{10}^{-12}F$

**Charge in each capacitor plate** $Q=3.90\mu C=3.9\times{10}^{-6}C$

**Part (a)**

As per the expression for **Charge on the Capacitor** $Q$:

\[Q=CV\]

\[V=\frac{Q}{C}\]

\[V=\frac{3.9\times{10}^{-6}C}{920\times{10}^{-12}F}\]

\[Potential\ Difference\ V=4239.13V\]

**Part (b)**

Given that the **Separation between the Capacitor Plates** $d$ is **doubled,** keeping the **charge **$Q$** constant**, so:

\[V_2=\frac{Q}{C_2}\]

As per the expression for **Capacitance of the Capacitor** $C$, if the **distance** $d$ is **doubled**:

\[C_2=\frac{\varepsilon_oA}{2d}=\frac{1}{2}(C)\]

Substituting in the above equation:

\[V_2=\frac{Q}{\dfrac{1}{2}(C)}\]

\[V_2=\frac{2Q}{C}\]

\[V_2=2V\]

\[V_2=\frac{2\times(3.9\times{10}^{-6}C)}{920\times{10}^{-12}F}\]

\[V_2=8478.26V\]

So the **Potential difference** $V$ is **doubled**, if the **separation between the capacitor plates** $d$ is **doubled**.

**Part (c)**

In order to calculate the amount of **work** $W$ that will be required to **double** the **separation between the capacitor plates**, we use the following expression:

\[W=\frac{1}{2}QV\]

By substituting the values in the above equation:

\[W=\frac{1}{2}(3.9\times{10}^{-6}C)\times(4239.13V)\]

\[W=8266.3\times{10}^{-6}J\]

\[Work\ Done\ W=0.008266.3J\]

## Numerical Result

**Part (a)** – The **Potential difference** $V$ existing between the plates of the capacitor is:

\[Potential\ Difference\ V=4239.13V\]

**Part (b)** – The **Potential difference** $V$ is **doubled** if the **separation between the capacitor plates** $d$ is **doubled**.

\[V_2\ =\ 2V=\ 8478.26\ V\]

**Part (c)** – The amount of **work** $W$ that will be required to **double** the **separation between the capacitor plates** $d$ will be:

\[Work\ Done\ W\ =\ 0.008266.3\ J\]

## Example

Calculate the **potential difference** $V$ across the **capacitor plates** if it has the **capacitance** of $245\ pF$ and the **electric charge** on each plate is $0.148\ \mu C$.

**Solution**

Given that:

**Capacitance of Capacitor** $C\ =\ 245pF\ =\ 245\times{10}^{-12}F$

**Charge in each capacitor plate** $Q\ =\ 0.148\mu C\ =\ 0.148\times{10}^{-6}C$

As per the expression for **Charge on the Capacitor** $Q$:

\[Q=CV\]

\[V=\frac{Q}{C}\]

\[V=\frac{0.148\times{10}^{-6}\ C}{245\times{10}^{-12}F}\]

\[Potential\ Difference\ V=604.08V\]