Question

Two point charges are located along the y-axis. One charge of 3 nC is at the origin and a second charge of 6 nC is at y=30 cm. a) Calculate the potential at y=60 cm. b) Calculate the potential at y=-60 cm. c) Repeat this problem assuming the 6 nC is replaced with a -6 nC charge.

Answer 1

Given,

q1 = 3nC ; at 0,0

q2 = 6nC ; at y = 30 = 0.3 m

a)y = 60 cm = 0.6 m

The potential at y = 60 cm will be the sum of the individual potentials due to the charges.

V = V1 + V2

V = k q1/r1 + k q2/r2

V = 9 x 10^9 x 10^-9 (3/0.6 + 6/0.3) = 225 V

Hence, V = 225 V

b)y = -60 cm

V = V1 + V2

V = k q1/r1 + k q2/r2

V = 9 x 10^9 x 10^-9 (3/0.6 + 6/0.9) = 105 V

Hence, V = 105 V

c) V = V1 + V2

V = k q1/r1 + k q2/r2

V = 9 x 10^9 x 10^-9 (3/0.6 - 6/0.3) = -135 V

Hence, V = -135 V

b)y = -60 cm

V = V1 + V2

V = k q1/r1 + k q2/r2

V = 9 x 10^9 x 10^-9 (3/0.6 - 6/0.9) = -15 V

Hence, V = -15 V