Question

Two point charges \( q_1 \) and \( q_2 \), are located at \( \vec{r_1} \) and \( \vec{r_2} \) respectively in an external electric field \( \vec{E} \). Obtain an expression for the potential energy of the system.

Hint:

When charges are placed in an external electric field, the total potential energy of the system includes both the energy due to the electric field and the interaction energy between the charges.

Answer 1

Potential Energy of Two Point Charges in an External Electric Field
Given
Two point charges \( q_1 \) and \( q_2 \) located at positions \( \vec{r}_1 \) and \( \vec{r}_2 \) respectively.
An external electric field \( \vec{E} \). Potential Energy Due to External Electric Field
The potential energy of a charge \( q \) in an external electric field \( \vec{E} \) is:
\[ U_{\text{ext}} = q \phi(\vec{r}) \] For the two charges:
\[ U_{\text{ext}} = q_1 \phi(\vec{r}_1) + q_2 \phi(\vec{r}_2) \] Potential Energy Due to Interaction Between Charges
The potential energy due to the interaction between \( q_1 \) and \( q_2 \) is: \[ U_{\text{int}} = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{|\vec{r}_1 - \vec{r}_2|} \] Total Potential Energy
The total potential energy \( U \) of the system is:
\[ U = U_{\text{ext}} + U_{\text{int}} \] Substituting the expressions:
\[ U = q_1 \phi(\vec{r}_1) + q_2 \phi(\vec{r}_2) + \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{|\vec{r}_1 - \vec{r}_2|} \] Final Expression The potential energy of the system is: \[ \boxed{U = q_1 \phi(\vec{r}_1) + q_2 \phi(\vec{r}_2) + \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{|\vec{r}_1 - \vec{r}_2|}} \]