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

The electrostatic potential energy \( U \) of a system of point charges is given by the formula: \[ U = \frac{1}{4 \pi \epsilon_0} \sum_{i<j} \frac{q_i q_j}{r_{ij}} \] where \( q_i \) and \( q_j \) are the point charges, and \( r_{ij} \) is the distance between them. The change in energy due to the external electric field \( \vec{E} \) is given by the work done by the electric field on the system, which is: \[ \Delta U = - \sum_{i} q_i \vec{E}. \vec{r_i} \] Here, the charges are placed at \( (-3 \, \text{cm}, 0, 0) \) and \( (3 \, \text{cm}, 0, 0) \), and the electric field \( \vec{E} = \frac{A}{r^2} \hat{r} \) is acting on them. The calculations will involve evaluating the work done on each charge and summing them up.