# physics

At what distance (in m) from a 3.8 mC charge is the electric potential equal to 32 V?

(k = 9 x 109 Nm2/C2)

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1. Let the distance be x. Solve the equation

kQ/x = V = 32 N*m/C,

with Q = 3.8*10^-3 C
V is the potential (referred to zero at infinite x)

x = kQ/32 = 9*10^9 (N*m^2/C^2)* (3.8*10^-3 C)/32 (N*m/C)
= 1.07*10^6 m

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