Question

an electric field exerts a force of 3.00 E -4 N on a positive test charge of 7.20 E-4 C. The magnitude of the field at this location of the charge is

a) 0.24 N/C
b) 0.417 N/C
c) 15.6 N/C
d) 21.6 N/C
e) 2.4 N/C

Answer 1

use the formula E = F/q where E is the magnitude of the field, F is force and q is positif test charge..,

Answer 2

are you find the answer?

Answer 3

or you could see his unit, if N/C implies N divided by C.., ok??