Asked by Erica
Four charges on a rectangle that is 6 cm by 3 cm. All but one of the corners has a positive charge. The magnitudes of the charges are all 8.40e-12 C. What is the net electric field at the center of the rectangle?
Answers
Answered by
drwls
The electric field contribution of the two opposite positive charges cancel out. The two charges on opposite sides that are of opposite sign will double the field due to a single charge. The center of the rectangle is at a distance R from corners that is given by the equation
R^2 = 3^2 + 1.5^2
R = 3.354 cm
Q = 8.40*10^-12 C
E = 2 * kQ^2/R^2
R^2 = 3^2 + 1.5^2
R = 3.354 cm
Q = 8.40*10^-12 C
E = 2 * kQ^2/R^2
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