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Force in an electric field.

A positive test charge of 9.0E-5 C is placed in an electric field of 35.0 N/C intensity. What is the strength of the force exerted on the test charge?

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q = 9.0 * 10^ -5 C (This has 5 significant figures), E = 35.0 N/C (This has 3 ...

Solution Summary

Force in an electric field is determined. The test charge exerted is determined.

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