# Answer to Question #23622 in Electric Circuits for Alexis

Question #23622

What is the magnitude of the electric field due to a 4.0 E 9 C charge at a point 0.020 away?

Expert's answer

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

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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