Two equal charges, each of 2.0 × 10^ -7 C , are kept fixed at a...

Let the two fixed charges be $q_1 and$q_2$, and the third charge be$q_3$.<br><br>Given$q_1 = q_2 = q_3 = q = 2.0 \times 10^{-7} \text{ C}$.<br><br>The initial position of$q_3$is midway between$q_1$and$q_2$. The distance from both charges is$r_{1i} = r_{2i} = 10 \text{ cm} = 0.1 \text{ m}$.<br><br>The initial potential energy of the third charge is:<br><br>$U_i = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_3}{r_{1i}} + \dfrac{1}{4\pi\epsilon_0} \dfrac{q_2 q_3}{r_{2i}}$<br><br>$U_i = 9 \times 10^9 \times \left(\dfrac{(2.0 \times 10^{-7})^2}{0.1} + \dfrac{(2.0 \times 10^{-7})^2}{0.1} \right)

U_i = 9 \times 10^9 \times \left(\dfrac{4 \times 10^{-14}}{0.1} \times2 \right) = 7.2 \times 10^{-3} \text{ J}

The final position of $q_3$ is $20 \text{ cm}$ from both charges. The distance from both charges is $r_{1f} = r_{2f} = 20 \text{ cm} = 0.2 \text{ m}$.

The final potential energy of the third charge is:

$U_f =$\dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_3}{r_{1f}} + \dfrac{1}{4\pi\epsilon_0} \dfrac{q_2 q_3}{r_{2f}}

$U_f = 9 \times 10^9 \times \left( \dfrac{(2.0 \times 10^{-7})^2}{0.2} + \dfrac{(2.0 \times 10^{-7})^2}{0.2} \right)$

$U_f = 9 \times 10^9 \times \left( \dfrac{4 \times 10^{-14}}{0.2} \times 2 \right) = 3.6 \times 10^{-3} \text{ J}$

The work done by the electric field is equal to the negative of the change in potential energy:

$W = -\Delta U = U_i - U_f$

$W = 7.2 \times 10^{-3} \text{ J} - 3.6 \times 10^{-3} \text{ J} = 3.6 \times 10^{-3} \text{ J}$