Question

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

Answer 1

Solution:-

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

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

• 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

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

• Potential of the point 12m from each charge

$\qquad\qquad \begin{aligned} \small V_2 &=\small (9\times10^9)\Bigg[\frac{35\times 10^-6}{0.12}+\frac{35\times 10^{-6}}{0.12}\Bigg]\\ &=\small 5.25\times10^6V \end{aligned}$

• Then the required work is

$\qquad\qquad \begin{aligned} \small W&=\small 0.50\times 10^{-6}C\times 5.25\times 10^6V\\ &=\small \bold{2.625J} \end{aligned}$

• The $\small 0.5\mu C$ charge moves perpendicular to both charges