Question

What is the magnitude (in N/C) and direction of an electric field that exerts a 3.50 ✕ 10−5 N upward force on a −1.55 µC charge?

Answer 1

Answer:

The magnitude of electric field is  22.58 N/C

Solution:

Given:

Force exerted in upward direction, $\vec{F_{up}} = 3.50\times 10^{- 5} N$

Charge, Q = $- 1.55\micro C = - 1.55\times 10^{- 6} C$

Now, we know by Coulomb's law,

$F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}$

Also,

Electric field, $E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}$

Thus from these two relations, we can deduce:

F = QE

Therefore, in the question:

$\vec{E} = \frac{\vec F_{up}}{Q}$

$\vec{E} = \frac{3.50\times 10^{- 5}}{- 1.55\times 10^{- 6}}$

$\vec{E} = - 22.58 N/C$

Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

$|\vec{E}| = 22.58\ N/C$