Question

When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What is the electric field at the origin?

Answer 1

Answer:

Electric field, $E=4\times 10^5\ N/C$

Explanation:

It is given that,

Magnitude of charge, $q_o=2\ nC=2\times 10^{-9}\ C$

Force experienced, $F=8\times 10^{-4}\ N$

We need to find the electric field at the origin. It is given by :

$F=q_o\times E$

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

$E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}$

$E=4\times 10^5\ N/C$

So, the electric field at the origin is $4\times 10^5\ N/C$. Hence, this is the required solution.