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 : Electric field, E = 0.417 N/C

Explanation :

It is given that,

Force due to charge,

Magnitude of charge,

We know that the electric field at a point is given by the force acting per unit charge.

So, the correct option is (b) " E = 0.417 N/C "

Hence, this is the required solution.

false

explanation:

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