# physics help please ?

A positive test charge of 7.45 × 10^−5 C is

placed in an electric field of 51.42 N/C intensity.

What is the strength of the force exerted on

the test charge?

Answer in units of N

### 1 Answer

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- az_lenderLv 74 weeks agoFavorite Answer
Super simple. F = Eq

= (51.42 N/C)(7.45 x 10^(-5) C)

= about 3.7 x 10^(-3) C but use a calculator.

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