Question

The electric potential at a position located a distance of 21.6 mm from a positive point charge of 7.80×10-9C and 14.2 mm from a second point charge is 1.30 kV. Calculate the value of the second charge.

Answer 1

According to the concept of the electric potential

Given that

Electric potential V=1.3*10^3 V

Charge q1=7.8*10^-9 C

Distance r1=21.6 mm

Distance r2=14.2 mm

Now we find the second charge

Basing on the concept of super position

V=K(q1/r1+q2/r2)

1.3*10^3=9*10^9{(7.8*10^-9/0.0216)+(q2/0.0142)}

1.3*10^-6/9=3.6*10^-7+q2/0.0142

0.14*10^-6=0.36*10^6+q2/0.0142

(0.14-0.36)*10^-6=q2/0.0142

-0.22*10^-6=q2/0.0142

Charge q2=(0.0142*-0.22*10^-6)=-0.003124*10^-6

=-3.1*10^-9 C