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 \( U \) between two point charges is defined by the formula:\[U = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r}\]Here, \( r \) represents the separation distance between the charges. Given \( q_1 = 5 \, \mu C \) and \( q_2 = -1 \, \mu C \), and a distance of 6 cm (derived from their positions at \( (-3 \, \text{cm}, 0, 0) \) and \( (3 \, \text{cm}, 0, 0) \)), the initial electrostatic energy is calculated as:\[U_{\text{initial}} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{5 \times 10^{-6} \times (-1 \times 10^{-6})}{0.06} = -\frac{5 \times 10^{-12}}{4 \pi \varepsilon_0 \cdot 0.06}\]The energy modification attributed to the electric field will be determined by the energy contained within that field, following the relation:\[U_{\text{field}} = \frac{1}{2} \varepsilon_0 E^2 V\]where \( E = \frac{A}{r^2} \) denotes the electric field strength and \( V \) is the volume influenced by the field. The alteration in electrostatic energy will be computed based on these parameters.