The electric potential at a position located a distance of 20.1 mm from a positive point...

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The electric potential at a position located a distance of 20.1 mm from a positive point...

The electric potential at a position located a distance of 20.1 mm from a positive point charge of 8.70×10^-9C and 13.9 mm from a second point charge is 1.26 kV. Calculate the value of the second charge.

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Answer 1

Answer #1

V = kq1 / r1 + kq2 / r2

1.26e3 = 9e9*8.70e-9 / 20.1e-3 + 9e9*q2 / 13.9e-3

-2635.522 = 6.4748e11 * q2

q2 = - 4.0704e-9 C

q2 = - 4.0704 nC

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Answer 2

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