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
- 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.