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

Answer : Electric field, E = 0.417 N/C

Explanation :

It is given that,

Force due to charge, $F=3\times 10^{-4}\ N$

Magnitude of charge, $q=7.2\times 10^{-4}\ C$

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

$E=\dfrac{F}{q}$

$E=\dfrac{3\times 10^{-4}\ N}{7.2\times 10^{-4}\ C}$

$E=0.417\ N/C$

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

Hence, this is the required solution.

Answer 2

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