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Question
A charge 6 µC is placed at the origin and another charge - 5 µC is placed on the y-axis at A = (0, 6.0 m).
(a) Calculate the net electric potential at P = (8.0 m, 0).
(b) Calculate the work done in bringing a proton from infinity to point P. What is the significance of the negative sign?
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Solution
q1= 6 µC = 6 × 10-6 C,
q2 = - 5 µC = -5 × 10-6 C
A ≡ (O, 6.0 m), P ≡ (8.0 m, 0), r1 = OP = 8 m,
q = e = 1.6 × 10-19 C, 1/4πε0 = 9 × 109 N m2/C2
`"r"_2 = "AP" = sqrt((8-0)^2 + (0 - 6)^2) = sqrt(64 + 36) = 10` m
(a) The net electric potential at P as a result of the two-charge system is
V = V1 + V2 = `1/(4piε_0)["q"_1/"r"_1 + "q"_2/"r"_2]`
`= (9 xx 10^9)[(6 xx 10^-6)/8 + (- 5 xx 10^-6)/10]`
`= (9 xx 10^3)(0.75 - 0.5)`
`= 2.25 xx 10^3 "V"`
= 2.25 kV
(b) The work done per unit charge by the electric field of the system of charges q1 and q2 in bringing a test charge from infinity to that point is the negative of the electric potential V at the point P.
V = `- "W"/"q"_0`
∴ W = - qV = - (1.6 × 10-19)(2.25 × 103)
= - 3.6 × 10-16 J
= - 2.25 keV
That is, the work done by the electric field on the positively charged proton in bringing it from a lower potential to a higher potential is negative, which means that an external agent must bring the proton against the electric field of the system of the two source charges.
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