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What is the strength of the electric field of a point charge of magnitude -4.8 × 10^-19 C at a distance of 4.0 × 10^-3 m? A.-3.6 × 10^-4 N/C B.-2.7 × 10^-4 N/C C.3.6 × 10^-4 N/C D.2.7 × 10^-4 N/C

Sagot :

Given:

The point charge has a magnitude

[tex]\begin{gathered} q=\text{ -4.8}\times10^{-19}\text{ C} \\ |q|=4.8\text{ }\times10^{-19\text{ }}C \end{gathered}[/tex]

The distance is

[tex]r=\text{ 4}\times10^{-3}\text{ m}[/tex]

To find the strength of the electric field.

Explanation:

The strength of the electric field can be calculated by the formula

[tex]E=\frac{k|q|}{r^2}[/tex]

Here, the constant is

[tex]k=9\times10^9\text{ Nm}^2\text{ /C}^2[/tex]

On substituting the values, the strength of the electric field will be

[tex]\begin{gathered} E=\frac{9\times10^9\times4.8\times10^{-19}}{(4\times10^{-3})^2} \\ =\text{ 2.7}\times10^{-4}\text{ N/C} \end{gathered}[/tex]

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