Answered

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

A0.24 N/C
B0.417 N/C
C15.6 N/C
D21.6 N/C
E2.4 N/C

Verified Answer :

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

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.

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