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Question: The force exerted by an electric field on charge of \( 5 \) micro-coulomb is \( 10 \times {10^{ - 4}...

The force exerted by an electric field on charge of 55 micro-coulomb is 10×10410 \times {10^{ - 4}} N. What is the electric intensity of the field at the point?

Explanation

Solution

The electric field intensity due to a positive charge is always directed away from the charge and the intensity due to a negative charge is always directed towards the charge.
It is denoted by E its SI unit is N/C.

Complete step by step answer:
A measure of the force exerted by one charged body on another. The electric field intensity at any location is the force that would be experienced by unit test charge placed at the location. Electric field intensity at any point is the strength of the electric field at that point. It is defined as the force experienced by unit positive charge placed at that point.
Electric field E=FqE = \dfrac{F}{q}
Or
E=10×1045×106=200N/CE = \dfrac{{10 \times {{10}^{ - 4}}}}{{5 \times {{10}^{ - 6}}}} = 200N/C

Additional Information:
Electric field is not scalar quantity because it depends upon the force, the direction of electric field is also the same as the direction of applied force. As force is vector quantity hence electric field intensity is also vector quantity.

Note:
The basic difference between electric field and electric field intensity is that the electric field is a region around a charge in which it exerts an electrostatic force on other charges. While the strength of the electric field at any point in space around an electric charge which its influence can be felt is known as the electric field. The electric field experienced by a unit positive charge placed at that.