Question

Two circular plates, each with a radius of 8.22 cm and equal and opposite charges of magnitude 7.952 μC, have an electric field between them. Calculate the electric field in N/C.

As the plates are slowly pulled apart, doubling the separation distance while keeping it small compared to the diameter of the plates, what changes occur with the electric field between the plates?
A. The electric field increases by a factor of 4.
B. The electric field stays the same.
C. The electric field decreases by a factor of 4.
D. The electric field increases by a factor of 2.
E. The electric field decreases by a factor of 2.

Answer 1

The electric field between the plates stays the same as they are pulled apart. So the orrect answer is Option B.

Step-by-step explanation:

The electric field between two parallel plates with equal and opposite charges is given by the formula E = V/d, where E is the electric field, V is the voltage difference between the plates, and d is the separation distance. In this case, the charges on the plates have a magnitude of 7.952 μC, the voltage difference between the plates is 0 (since they are connected), and the separation distance is doubled. As a result, the electric field between the plates stays the same. Therefore, the correct answer is B. The electric field stays the same.