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The electric potential at a position located a distance of 20.1 mm from a positive point charge of 6.50×10-9C and 12.1 mm from a second point charge is 1.26 kV. Calculate the value of the second charge.

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potential is the scalar sum...
1.26E3=k6.5E-9/.0201 + kQ/.0121

solve for Q

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