A proton (charge q = 1.6 x 10-19 C) is placed in an electric field of...

Question

Question

A proton (charge q = 1.6 x 10-19 C) is placed in an electric field of magnitude 400 N/C. What is the magnitude of the force o

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Answer 1

Answer #1

Answer: Force due to electric field is given as : F = qE

where, q = charge = 1.6*10^-19 C and E - electric field = 400 N/C

=> F = 1.6*10^-19 * 400

=> F = 6.4*10^-17 N. Answer

Hence, option (d) is correct.

Answer 2

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