Question

The electric potential at a position located a distance of 20.7 mm from a positive point charge of 8.60×10-9C and 15.1 mm from a second point charge is 1.14 kV. Calculate the value of the second charge.

Answer 1

Answer:

q2 = -4.35*10^-9C

Explanation:

In order to find the values of the second charge, you use the following formula:

$V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}$       (1)

V: electric potential = 1.14 kV = 1.14*10^3 kV

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1: charge 1 = 8.60*10^-9 C

q2: charge 2 = ?

r1: distance to the first charge = 20.7mm = 20.7*10^-3 m

r2: distance to the second charge = 15.1mm

You solve the equation (1) for q2, and replace the values of the other parameters:

$q_2=\frac{r_2}{k}[V-k\frac{q_1}{r_1}]=\frac{Vr_2}{k}-\frac{q_1r_2}{r_1}\\\\q_2=\frac{(1.14*10^3V)(15.1*10^{-3}m)}{8.98*10^9Nm^2/C^2}-\frac{(8.60*10^{-9}C)(15.1*10^{-3}m)}{20.7*10^{-3}m}\\\\q_2=-4.35*10^{-9}C$

The values of the second charge is -4.35*10^-9C