What is the magnitude of the electric field due to a 4.0 E 9 C charge at a point 0.020 away?
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Expert's answer
2013-02-12T09:12:34-0500
The formula for the magnitudeof the electric field
E = q / (4*pi*eps0*R^2), where q is the charge, eps0 is the permittivity of vacuum and R is the distance from the charge. Assuming q = 4.0*10^(-9) C and R = 0.020 m,
E = 4.0*10^(-9) C / (4*pi*8.85 * (10^(-12) F/m) * (0.020 m)^2) = 8992 V / cm
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