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STALIN [3.7K]
3 years ago
15

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?

Physics
1 answer:
IRISSAK [1]3 years ago
6 0

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

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