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

Question

1 month ago

11

is the electric field at the origin?

1 answer:

Softa [3K] · Answer 1 · 2026-04-06T20:53:29+03:00

Softa [3K]1 month ago

3 0

Answer:

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

Explanation:

The following values are provided:

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

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

To determine the electric field at the origin, the formula employed is:

$F=q_o\times E$

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

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

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

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