Question

Two parallel plates have equal but opposite charges on their surface. The plates are separated by a finite distance.

Answer 1

Answer:

(a) The speed of proton decreases as it moves from A to B.

(b) Plate B is at a higher potential.

(c) Plate B is positive, plate A is negative.

(d) Parallel lines parallel to the two plates.

    Parallel lines equally spaced.

Explanation:

The electric potential energy is given by the following formula:

$U = \frac{1}{4\pi \epsilon_0}\frac{qq_0}{r}$

Alternatively, potential energy in a uniform electric field is

$U = qEr$

where 'r' is the distance from negative to positive plates. This definition is analogues to that of gravitational potential energy, U = mgh.

If the positively charged proton is gaining potential energy as it gets closer to plate B, then plate B is charged positively.

(a) According to this information, the speed of proton decreases as it moves from A to B. This is similar to the speed of an object which is gaining potential energy by moving higher.

(b) By the same gravitational analogy, plate B is at a higher potential.

(c) As explained before, Plate A is negative and Plate B is positive.

(d) The equipotential lines are parallel to electric field lines which are perpendicular to the plates. So, the equipotential lines are parallel to the plates. Since the electric field between the plates is uniform, then the equipotential lines are equally seperated.