# What is the magnitude of a point charge that would create an electric field of 4.80 N/C at points...

## Question:

What is the magnitude of a point charge that would create an electric field of 4.80 N/C at points 2.90 m away?

## Electric Field Due to a Point Charge

Electric field at a point is the Coulombic force experienced by a unit positive charge placed at that point. It is a vector quantity and obeys superposition principle for multiple point charges. The electric field due to a point charge depends on its position with respect to the point charge.

## Answer and Explanation:

The electric field due to a point charge is given by,

{eq}\displaystyle{ E=\frac{kq}{r^2} } {/eq}

Where,

- {eq}k=8.99\times10^9\ Nm^2/C^2 {/eq} is the Coulombic constant,

- {eq}q {/eq} is the magnitude of charge and

- {eq}r {/eq} is the distance from the charge.

Given:

{eq}E=4.80\ N/C\\ r=2.90\ m {/eq}

The magnitude of charge is ,

{eq}\displaystyle{ q=\frac{Er^2}{k}=\frac{4.80\times2.90^2}{8.99\times10^9}\ C=4.49\times10^{-9}\ C } {/eq}

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