Question

A point charge with charge q1 = 3.10 μC is held stationary at the origin. A second point charge with charge q2 = -4.50 μC moves from the point ( 0.140 m , 0) to the point ( 0.265 m , 0.265 m ). How much work W is done by the electric force on the moving point charge?

Answer 1

Answer:

W = 0.562 J

Explanation:

Initial potential energy of two point charges is given as

$U_i = \frac{kq_1q_2}{r_1}$

here we have

$q_1 = 3.10 \mu C$

$q_2 = -4.50 \mu C$

$r_1 = 0.140 m$

now we have

$U_i = \frac{(9\times 10^9)(3.10 \mu C)(-4.50 \mu C)}{0.140}$

$U_i = -0.897 J$

now at final position the distance between two charges is given as

$r_2 = \sqrt{0.265^2 + 0.265^2} = 0.375 m$

Now final energy is given as

$U_f =\frac{kq_1q_2}{r_2}$

$U_f = \frac{(9\times 10^9)(3.10 \mu C)(-4.50 \mu C)}{0.375}$

$U_f = -0.335 J$

Now the work done to change the position of charge is given as

$W = U_f - U_i$

$W = (-0.335) - (-0.897) = 0.562 J$