Answer:
Explanation:
The capacitance of a capacitor in terms of the dielectric constant, area of the plate and the distance separating the plate is given by:
Where A = Area of the plate
d = distance between the plates
dielectric constant
Case 1:
When a meta slab of thickness, a, is added between the plates of the parallel plate capacitor , the effective separation between the plates becomes d+a
Therefore the capacitance of the capacitor becomes:
.......................(1)
Case 2:
Introducing a dielectric with dielectric constant K between the plates, the capacitance of the capacitor becomes:
.........................(2)
Equating (1) and (2)
The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.
<h3>How does test charge affect electric field?</h3>
As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.
Adjusting the amount of charge on the test charge will not change the electric field force.
<h3>What is a test charge used for?</h3>
The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.
To learn more about test charge, refer
brainly.com/question/16737526
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