Question

A proton is moved 15 cm on a path parallel to the field lines of a uniform electric field of $ 2.0 \times {10^5}V/{m^3} $ . What are the possible changes in potential? Consider both cases of a moving proton.

Answer 1

Hint: the two possible ways the proton can be moved parallel to the electric field is towards the source of the field or away from the source of the field. The change in potential is equal to the negative to the product of the electric field and the displacement of the charge.

Formula used: In this solution we will be using the following formulae;
 $ V = - Ed $ where $ V $ is the change in potential or the potential difference between the two points, $ E $ is the electric field strength and $ d $ is the displacement of the charge.

Complete Step-by-Step solution:
Generally, when a charge moves parallel to an electric field, the potential changes. The change in potential is equal to the negative of the electric field strength and the displacement of the charge.
This is mathematically given as
 $ V = - Ed $ where $ V $ is the change in potential or the potential difference between the two points, $ E $ is the electric field strength and $ d $ is the displacement of the charge.
Hence, from the given values, we have, for proton moving away from the charge, that
 $ V = - 2.0 \times {10^5} \times \left( {0.15} \right) = - 3.0 \times {10^4}V $
And for proton moving towards the source, we have
 $ V = - 2.0 \times {10^5} \times \left( { - 0.15} \right) = 3.0 \times {10^4}V $
Those are the two possible changes in potential.

Note:
The negative in potential makes the change in potential positive as when change potential energy is positive. For example, potential energy of a proton is positive when the proton moves towards the source of the field. The displacement in this case will be negative, and so, the negative in the displacement multiplies the negative in the formula, to make it positive.