Question

A + 35 μC point charge is placed 32 cm from an identical + 35 μC charge. How much work would be required to move a + 0.50μC test charge from a point midway between them to a point 12cm closer to either of the charges?

Answer 1

Explanations & Calculations

• Work is the energy needed to move between the two mentioned points and it is given by

$\qquad\qquad \begin{aligned} \small W&=\small q\Delta V \end{aligned}$

• Now q is known & the potential difference between those points are to found to complete the math.
• Potential of the point midway between the like charges is

$\qquad\qquad \begin{aligned} \small V_1&=\small k\frac{q_1}{r_1}+k\frac{q_2}{r_2}\\ &=\small (9\times 10^9Nm^2C^-2 )\Bigg[\frac{35\times 10^{-6}C}{0.16m}+\frac{35\times10^{-6} C}{0.16m}\Bigg]\\ &=\small 3.9375\times 10^6 V \end{aligned}$

• When moved 12cm closer to either charge, one makes 4cm with the 0.5uc and the other makes (16cm+12cm=) 28cm with it. Then the potential of the point 12m from either charge,

$\qquad\qquad \begin{aligned} \small V_2 &=\small (9\times10^9)\Bigg[\frac{35\times 10^-6}{0.04m}+\frac{35\times 10^{-6}}{0.28m}\Bigg]\\ &=\small 9.00\times10^6V \end{aligned}$

• Then the required work is

$\qquad\qquad \begin{aligned} \small W&=\small 0.50\times 10^{-6}C\times (9.00-3.9375)\times10^6V\\ &=\small \bold{2.53\,J} \end{aligned}$