Question

A 5 n force is acting on the charge 6μc at any point. what is magnitude of the electric field at that point?

Answer 1

The magnitude of the electric field at the given point is 8.33 * 10⁵N/C.

Given the data in the question;

• Electrostatic force exterted; $F = 5N$

• Charge; $q = 6uC = 6*10^{-6}C$

• Magnitude of electric field; $E =\ ?$

Electric Field

The magnitude of electric field is simply referred to as the force per charge on a test charge.

It is expressed as;

$E = \frac{F}{q}$

Here, F is electrostatic force exerted and q is the charge.

To determine the magnitude of the electric field at the given point, we substitute our given values into the expression above.

$E = \frac{F}{q} \\\\E = \frac{5N}{6*10^{-6}C} \\\\E = 8.33 * 10^5N/C$

Therefore, the magnitude of the electric field at the given point is 8.33 * 10⁵N/C.

Learn more about coulomb's law: brainly.com/question/506926