Question

two circular plates, each with a radius of 8.22 cm, have equal and opposite charges of magnitude 3.052 μc. calculate the electric field between the two plates. assume that the separation distance is small in comparison to the diameter of the plates. electric field: n/c the plates are slowly pulled apart, doubling the separation distance. again, assume the separation distance remains small in comparison to the diameter of the plates. what changes occur with the electric field between the plates? the electric field decreases by a factor of 2. the electric field stays the same. the electric field increases by a factor of 2. the electric field decreases by a factor of 4. the electric field increases by a factor of 4.

Answer 1

If the separation distance is doubled, then the electric field decreases by a factor of 4.

What is the electric field strength?

We know that the electric field strength is known to depend on the magnitude of the charge and the distance of separation. We know that the electric field refers to the region in which the influence of a charge is felt. Recall that a charge is a specie that is positively or negatively charged. The charge on a specie must always be shown by its sign.

We know that the electric field is the region in space where the influence of a charge can be felt. If a charge is placed in the vicinity of another charge, the second charge would experience a force due to the presence of the first charge. This is because the second charge was brought into the electric field of the first charge.

Thus we know that;

E = Kq/r^2

Where;

E = electric field strength

q = magnitude of charge

r = distance of separation

Now;

E = 9.0* 10^9 * 3.052 * 10^-6/(8.22 * 10^-2)^2

E = 4 N/C

Given that the electric filed strength is inversely proportional to the distance of separation, when the distance between the charges is doubled, the electric field decreases by a factor of 4.

Learn more about electric field strength:brainly.com/question/15170044?

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