Question

A charge of +15nC lies on the x-axis at x=-15cm, while a charge of -30nC lies on the x-axis at x=+15cm. What is the voltage at the origin? What is the change in potential energy of a +10nC charge that moves from the origin along the y-axis to y=+15cm?

Answer 1

here,

the charge Q1 = 15 nC = 15 * 10^-9 C is at x1 = - 15 cm = - 0.15 m

the charge Q2 = - 30 nC = - 30 * 10^-9 C is at x2 = 15 cm = 0.15 m

the voltage at origin , V = K * Q1 /x1 + K * Q2 /x2

V = 9 * 10^9 * 10^-9 /0.15 * ( 15 - 30) V

V = 900 V

Q3 = 10 nC = 10 * 10^-9 C

y = 15 cm = 0.15 m

the change in potential energy of a +10nC charge that moves from the origin along the y-axis to y = +15cm , dU = Uf - Ui

dU = Q3 * ((K * Q1 /sqrt(x1^2 + y^2) + (K * Q1 /sqrt(x1^2 + y^2) - ( (K * Q1 /(x1)) + (K * Q2 /(x2))

dU = 9 * 10^9 * 10 * 10^-9 * 10^-9 * ( (15 /sqrt(0.15^2 + 0.15^2) - 30 /sqrt(0.15^2 + 0.15^2)) - (15 /0.15 - 30 /0.15)) J

dU = 2.64 * 10^-6 J