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

Question

Sullivan Carrillo

Physics

28 February, 05:28

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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Elliott Huffman · 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.