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

3.00 E-4 / 7.20 E-4 = 0.417
answer is b

Answer 2

Answer : Electric field, E = 0.417 N/C

Explanation :

It is given that,

Force due to charge,
`F=3* 10^(-4)\ N`

Magnitude of charge,
`q=7.2* 10^(-4)\ C`

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

`E=(F)/(q)`

`E=(3* 10^(-4)\ N)/(7.2* 10^(-4)\ C)`

`E=0.417\ N/C`

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

Hence, this is the required solution.