Question

A point charge${\mathit{q}}_{\mathbf{1}}$is held stationary at the origin. A second chargeis placed at point, and the electric potential energy of the pair of charges is$\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{J}$. When the second charge is moved to point b, the electric force on the charge does$\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{J}$work. What is the electric potential energy of the pair of charges when the second charge is at point?

Short answer

The electric potential energy of the pair of charges when the second charge is at point bis $+7.3\times {10}^{-8}\mathrm{J}$.

Step-by-step solution

Definition of Potential energy

Consider the formula for the work required to move a charge from the point a to the point b is:

${\mathbf{W}}_{\mathbf{a}\mathbf{\to }\mathbf{b}}\mathbf{=}{\mathbf{U}}_{\mathbf{a}}\mathbf{-}{\mathbf{U}}_{\mathbf{b}}$

Here, $U_a$and $U_b$are the initial and final electric potential respectively.

Determine the electric potential energy of the pair of charges when the second charge is at point

Consider the given initial electric potential as${\mathrm{U}}_{\mathrm{a}}=+5.4\times {10}^{-8}\hspace{0.33em}\mathrm{J}$.

Chargemoves to pointwhere work done is${\mathrm{W}}_{\mathrm{a}\to \mathrm{b}}=-1.9\times {10}^{-8}\hspace{0.33em}\mathrm{J}$.

Substitute the values in the equation (1) and solve as:

$-1.9\times {10}^{-8}\hspace{0.33em}\mathrm{J}=+5.4\times {10}^{-8}\hspace{0.33em}\mathrm{J}-{\mathrm{U}}_{\mathrm{B}}$

Solve for$U_B$as:

$$\begin{gathered} {\mathrm{u}}_{\mathrm{b}}=\left(+5.4\times {10}^{-8}\mathrm{J}\right)-\left(-1.9\times {10}^{-8}\mathrm{J}\right) \\ =+7.3\times {10}^{-8}\mathrm{J} \end{gathered}$$

Hence, the electric potential energy of the pair of charges when the second charge is at point bis $+7.3\times {10}^{-8}\hspace{0.33em}\mathrm{J}$.