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3 years ago

What magnitude charge creates a 4.70 N/C electric field at a point 3.70 m away?

Physics

1 answer:

Lana71 [14]3 years ago

6 0

Answer:

The magnitude of charge is 7.15 nC.

Explanation:

Given that,

Electric field = 4.70 N/C

Distance = 3.70 m

We need to calculate the magnitude of charge

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$q = \dfrac{Er^2}{k}$

Where, k = Boltzmann constant

r = distance

q = charge

E = electric field

Put the value into the formula

$q =\dfrac{4.70\times(3.70)^2}{9\times10^{9}}$

$q=7.15\times10^{-9}\ C$

$q = 7.15\ nC$

Hence, The magnitude of charge is 7.15 nC.

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Answer:

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