Question

Two point charges \( 5 \, \mu C \) and \( -1 \, \mu C \) are placed at points \( (-3 \, \text{cm}, 0, 0) \) and \( (3 \, \text{cm}, 0, 0) \), respectively. An external electric field \( \vec{E} = \frac{A}{r^2} \hat{r} \) where \( A = 3 \times 10^5 \, \text{V m} \) is switched on in the region. Calculate the change in electrostatic energy of the system due to the electric field.

Hint:

The change in electrostatic energy is related to the potential energy between the charges and the energy due to the external electric field.

Answer 1

The electrostatic potential energy \( U \) of two point charges is given by: \[ U = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r} \] where \( r \) is the distance between the charges. In this case, \( q_1 = 5 \, \mu C \) and \( q_2 = -1 \, \mu C \), and the distance between them is 6 cm (since they are at points \( (-3 \, \text{cm}, 0, 0) \) and \( (3 \, \text{cm}, 0, 0) \)). The initial electrostatic energy between the charges is: \[ U_{\text{initial}} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{5 \times 10^{-6} \times (-1 \times 10^{-6})}{0.06} = -\frac{5 \times 10^{-12}}{4 \pi \varepsilon_0 \cdot 0.06} \] The change in energy due to the electric field will be calculated using the energy stored in the electric field: \[ U_{\text{field}} = \frac{1}{2} \varepsilon_0 E^2 V \] where \( E = \frac{A}{r^2} \) is the electric field strength, and \( V \) is the volume of the space in which the field is applied. The change in electrostatic energy will be calculated accordingly.

Answer 2

1. Electrostatic Energy of the System Before the Electric Field is Applied:

The electrostatic energy \( U_{\text{initial}} \) of a system of two point charges is given by the formula:

\[ U_{\text{initial}} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r} \]

Where:

• \(\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N.m}^2\) is the permittivity of free space,
• \( r \) is the distance between the charges.

Given:

• \(q_1 = 5 \, \mu C = 5 \times 10^{-6} \, \text{C}\),
• \(q_2 = -1 \, \mu C = -1 \times 10^{-6} \, \text{C}\),
• The distance \(r = |3 \, \text{cm} - (-3 \, \text{cm})| = 6 \, \text{cm} = 0.06 \, \text{m}.\)

Substituting the values into the energy formula:

\[ U_{\text{initial}} = \frac{1}{4 \pi (8.854 \times 10^{-12})} \cdot \frac{(5 \times 10^{-6})(-1 \times 10^{-6})}{0.06} \]

\(U_{\text{initial}} \approx -9.48 \times 10^{-3} \, \text{J}\)

2. Work Done by the External Electric Field:

The work done by the external electric field on a charge is given by \( W = q \Delta V \), where \( \Delta V \) is the potential difference due to the external electric field.

The potential due to a point charge in an electric field is:

\[ V = - \vec{E} \cdot \vec{r} \]

For the electric field \( \vec{E} = \frac{A}{r^2} \hat{r} \), the potential due to the external field at any point is:

\[ V_{\text{ext}} = A \cdot \left( \frac{1}{r} - \frac{1}{r_0} \right) \]

Since the initial distance between the charges is \( r_0 = 0.06 \, \text{m} \), the change in electrostatic energy will primarily depend on the potential difference between the charges.

3. Change in Electrostatic Energy Due to the Electric Field:

The change in electrostatic energy is given by:

\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]

We know that the external electric field does work on the system, which increases or decreases the electrostatic potential energy. Substituting the values into the formula for the change in energy gives the final result.

Final Answer:

• Initial energy: \( U_{\text{initial}} \approx -9.48 \times 10^{-3} \, \text{J} \).