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:

When an external electric field is applied, the change in electrostatic energy can be calculated by finding the work done by the electric field on the system of charges.

Answer 1

Change in Electrostatic Energy of a System due to an External Electric Field 

Given:

• Two point charges: \( q_1 = 5 \, \mu C = 5 \times 10^{-6} \, \text{C} \) and \( q_2 = -1 \, \mu C = -1 \times 10^{-6} \, \text{C} \)
• Positions of the charges: \( (-3 \, \text{cm}, 0, 0) \) and \( (3 \, \text{cm}, 0, 0) \) respectively
• External electric field: \( \vec{E} = \frac{A}{r^2} \hat{r} \), where \( A = 3 \times 10^5 \, \text{V/m} \) and \( r \) is the distance from the origin

Concepts:

The change in electrostatic energy is calculated by the work done by the external electric field on the system. The energy change in a system of charges due to an external electric field \( \vec{E} \) is given by: \[ \Delta U = \sum_i q_i \vec{E} \cdot \vec{r}_i \] where \( \vec{r}_i \) is the position vector of the charge \( q_i \).

1. Calculating the Work Done by the External Electric Field:

The external electric field is radial, directed along the position vectors of the charges, and its magnitude is given by \( E = \frac{A}{r^2} \).

Position of the charges:

• Charge \( q_1 = 5 \, \mu C \) is located at \( \vec{r_1} = (-3 \, \text{cm}, 0, 0) \), so \( r_1 = 3 \, \text{cm} = 0.03 \, \text{m} \)
• Charge \( q_2 = -1 \, \mu C \) is located at \( \vec{r_2} = (3 \, \text{cm}, 0, 0) \), so \( r_2 = 3 \, \text{cm} = 0.03 \, \text{m} \)

Force on each charge due to the external field:

The electric field at each position is: \[ E_1 = \frac{A}{r_1^2} = \frac{3 \times 10^5}{(0.03)^2} \, \text{V/m} = 3.33 \times 10^7 \, \text{V/m} \] and similarly, \[ E_2 = \frac{A}{r_2^2} = 3.33 \times 10^7 \, \text{V/m} \]

2. Work Done by External Electric Field:

The work done by the external electric field is: \[ W_1 = q_1 E_1 r_1 = (5 \times 10^{-6}) \times (3.33 \times 10^7) \times (0.03) \] \[ W_1 = 4.995 \, \text{J} \] Similarly, for \( q_2 \): \[ W_2 = q_2 E_2 r_2 = (-1 \times 10^{-6}) \times (3.33 \times 10^7) \times (0.03) \] \[ W_2 = -0.999 \, \text{J} \]

3. Total Work Done:

The total work done by the external electric field on the system is the sum of the individual works: \[ \Delta U = W_1 + W_2 = 4.995 + (-0.999) = 3.996 \, \text{J} \]

✔ Final Answer:

The change in the electrostatic energy of the system due to the external electric field is \( \boxed{3.996 \, \text{J}} \).