Question

What is the magnitude (in N/C) and direction of an electric field that exerts a 4.00 x 10^−5 N upward force on a −2.00 µC charge? magnitude N/C
direction: upward, downward, to the left, to the right

Answer 1

Answer:

The magnitude of electric field is $1.25\times10^{14}\ N/C$ in downward.

Explanation:

Given that,

Force $F= 4.00\times10^{-5}\ N$

Charge q= -2.00 μC

We know that,

Charge is negative, then the electric field in the opposite direction of the exerted force.

We need to calculate the magnitude of electric field

Using formula of electric force

$F = qE$

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

$E=\dfrac{4.00\times10^{-5}}{-2.00\times1.6\times10^{-19}}$

$E=-1.25\times10^{14}\ N/C$

Negative sign shows the opposite direction of electric force.

Hence, The magnitude of electric field is $1.25\times10^{14}\ N/C$ in downward.