Asked by tyler
A force of 0.053 N is required to move a charge of 36 µ a distance of 30 cm in a uniform electric field. What is the size of the electric potential difference between the two points?
Answers
Answered by
Damon
F * d = q * V
V = F * d / q
= .053 * .3 /36*10^-6
= 5.3 * 10^-2 * 3 * 10^-1 *10^6/36
= .442 * 10^3
= 442 volts
V = F * d / q
= .053 * .3 /36*10^-6
= 5.3 * 10^-2 * 3 * 10^-1 *10^6/36
= .442 * 10^3
= 442 volts
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