Question

A point charge q1=+5.00nC is at the fixed position x=0, y=0, z=0. You find that you must do 8.10×10−6J of work to bring a second point charge from infinity to the position x=+4.00cm, y=0, z=0. What is the value of the second charge?

Answer 1

The value of the second charge is 1.2 nC.

Electric potential

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

Magnitude of second charge

The magnitude of the second charge is determined by applying Coulomb's law.

$E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 = \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC$

Thus, the  value of the second charge is 1.2 nC.

Learn more about electric potential here: brainly.com/question/14306881