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

q2 = -4.35*10^-9C

Step-by-step explanation:

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

`V=k(q_1)/(r_1)+k(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=(r_2)/(k)[V-k(q_1)/(r_1)]=(Vr_2)/(k)-(q_1r_2)/(r_1)\\\\q_2=((1.14*10^3V)(15.1*10^(-3)m))/(8.98*10^9Nm^2/C^2)-((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