Chapter 4: Q2E (page 777)

A point charge $q_{1}$ is held stationary at the origin. A second chargeis placed at point, and the electric potential energy of the pair of charges is $+ 5.4 \times 10^{- 8} J$ . When the second charge is moved to point b, the electric force on the charge does $- 1.9 \times 10^{- 8} J$ work. What is the electric potential energy of the pair of charges when the second charge is at point?

Expert verified

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

Step by step solution

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

$W_{a \to b} = U_{a} - U_{b}$

Here, $U_{a}$ and $U_{b}$ are the initial and final electric potential respectively.

Consider the given initial electric potential as $U_{a} = + 5.4 \times 10^{- 8} J$ .

Chargemoves to pointwhere work done is $W_{a \to b} = - 1.9 \times 10^{- 8} J$ .

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

$- 1.9 \times 10^{- 8} J = + 5.4 \times 10^{- 8} J - U_{B}$

Solve for $U_{B}$ as:

$u_{b} = (+ 5.4 \times 10^{- 8} J) - (- 1.9 \times 10^{- 8} J) = + 7.3 \times 10^{- 8} J$

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

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