A point charge of $10^{-8}$ C is placed at origin. The work done in moving a point charge $2\ \mu C$ from point $A(4,4,2)$ m to $B(2,2,1)$ m is $\_\_\_\_$ J. $\left(\frac{1}{4\pi\epsilon_0}=9 imes10^9 ight.$ in SI units)

Question

A point charge of $10^{-8}$ C is placed at origin. The work done in moving a point charge $2\ \mu C$ from point $A(4,4,2)$ m to $B(2,2,1)$ m is $\_\_\_\_$ J. $\left(\frac{1}{4\pi\epsilon_0}=9	imes10^9ight.$ in SI units)

Accepted Answer

A. 1. The work done in moving a charge equals the charge times the potential difference between the final and initial points, so first find the distances of points A and B from the origin.
Given point A(4,4,2),
$$
r_A = \sqrt{4^2+4^2+2^2} = \sqrt{36} = 6\ 	ext{m}
$$
Given point B(2,2,1),
$$
r_B = \sqrt{2^2+2^2+1^2} = \sqrt{9} = 3\ 	ext{m}
$$

2. Compute the potential at each point due to the source charge at the origin.
Given $Q=10^{-8}$ C,
$$
V_A = \frac{kQ}{r_A} = \frac{(9	imes10^9)(10^{-8})}{6} = 15\ 	ext{V}
$$
$$
V_B = \frac{kQ}{r_B} = \frac{(9	imes10^9)(10^{-8})}{3} = 30\ 	ext{V}
$$

3. Use $W=q(V_B-V_A)$ to find the work done in moving the test charge from A to B.
Given $q=2	imes10^{-6}$ C,
$$
W = (2	imes10^{-6})(30-15) = 30	imes10^{-6}\ 	ext{J}
$$

Hence, the work done is $30	imes10^{-6}$ J, so the answer is option A.