A positive test charge of 5.00 e-5 c is placed in an electric field. the force on it is 0.751 n. the magnitude of the electric field at the location of the test charge is

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Physics

Adan

28 November, 09:56

A positive test charge of 5.00 e-5 c is placed in an electric field. the force on it is 0.751 n. the magnitude of the electric field at the location of the test charge is

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Felipe · Answer 1

In order to get the magnitude of the electric field at the location of the test charge, you have to use the formula: E=Fe/q.

Given E as electric field strength; Fe as electric force = 0.751 N and q as charge = 5*10-5 C. Therefore,

E = 0.751 N / 5 * 10-5 CE=15020 N C-1 or V m-1 Electric field at the location.