Question

An electric field of 244,000 N/C points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of -6.95 µC at this spot?

magnitude N

direction

Answer 1

Let west direction be -x.

So,

E = 244000 N/C(-)

Magnitude of Charge: q

9= -6.95 * 10-6 C

We know that Electric Force and Electric field are related by the below equation:

F = q*E

Substituting values in above equation:

F = -6.95 * 10-6* 244000 (-1)

F = 6.95 * 10-6 * 244000 ()

F = 1.6958 (î) N

Hence, Magnitude of Force = 1.6958 N

Direction of Force = +x = East direction

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