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.