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

Modification in Electrostatic Energy of a Configuration Under an External Electric Field

Provided Data:

• Charge \( q_1 = 5 \, \mu C = 5 \times 10^{-6} \, \text{C} \)
• Charge \( q_2 = -1 \, \mu C = -1 \times 10^{-6} \, \text{C} \)
• Position vectors: \( \vec{r}_1 = (-3 \, \text{cm}, 0, 0) \) for \( q_1 \) and \( \vec{r}_2 = (3 \, \text{cm}, 0, 0) \) for \( q_2 \)
• External electric field: \( \vec{E} = \frac{A}{r^2} \hat{r} \), with \( A = 3 \times 10^5 \, \text{V/m} \) and \( r \) being the radial distance from the origin.

Underlying Principles:

The alteration in electrostatic potential energy is equivalent to the work performed by the external electric field on the charged system. The general formula for the change in energy (\(\Delta U\)) of a system of charges (\(q_i\)) placed at positions (\(\vec{r}_i\)) within an external electric field (\(\vec{E}\)) is: \[ \Delta U = \sum_i q_i \vec{E} \cdot \vec{r}_i \]

1. Calculation of Work Done by External Field:

The external electric field is radial, aligned with the position vectors of the charges, and its strength is \( E = \frac{A}{r^2} \).

Charge Positions and Distances:

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

Force on Each Charge Due to External Field:

The electric field strength at each location is: \[ E_1 = \frac{A}{r_1^2} = \frac{3 \times 10^5}{(0.03)^2} \, \text{V/m} = 3.33 \times 10^7 \, \text{V/m} \] The field strength at the second charge's position is identical: \[ E_2 = \frac{A}{r_2^2} = 3.33 \times 10^7 \, \text{V/m} \]

2. Work Done by the External Field on Individual Charges:

The work done by the external field on \(q_1\) 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 on the System:

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

✔ Final Result:

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