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What is the magnitude of the electric field at the location of the test charge?

A positive test charge of 5.0 x 10^-4 C is in an electric field that exerts a force of 2.5 x 10^-4 N on it. What is the magnitude of the electric field at the location of the test charge?

Public Comments

1. By definition the force on the charge exerted by the electric field is: F = E*Q, were: F - force (N), E - electric field (V/m), and Q - electric charge (As). Therefore in your case: E = F/Q = 2.5*10^-4/5*10^-4 = 0.5 V/m.