Question

A charge of 5C experiences a force of $5000N$when it is kept in a uniform electric field. What is the potential difference between two points separated by a distance of $1cm$
A) $10V$
B) $250V$
C) $1000V$
D) $2500V$

Answer 1

Hint: Recall the concept of electric field and intensity. The space around an electric charge where its effect can be felt is known as electric field. The electric field intensity is the force experienced by a unit positive charge placed at that point.

Complete step by step solution:
Step I:
Given that the charge is $q = 5C$
Force experienced is $F = 5000N$
Distance between two points is $ = 1cm$
Or $0.01m$
Potential difference $V = ?$

Step II:
The formula for electric field is written as
$E = \dfrac{F}{q}$
$E = \dfrac{{5000}}{5}$
$E = 1000V$

Step III:
If there is a potential difference, then the electric field will also be there. Electric field is defined as the negative space derivative of electric potential. Also the relation between the electric field intensity and the potential can be written as
$E = - \dfrac{{dv}}{{dr}}$
Where E is the electric field
dv is the potential difference
dr is the path or the distance
Or $dv = E.dr$
Substituting the values and solving for the value of potential difference
$dv = 1000 \times 0.01$$dv = 10V$

Step IV:
$\therefore $the potential difference between two points when it is placed in a uniform electric field and separated by a distance of $1cm$is $10V$.
$ \Rightarrow $Option A is the right answer.

Note: It is important to note that if the charge is uniform at all the points, then the electric field is zero. The presence of potential difference does not affect the electric field. If the electric field is directed from a lower potential to a higher potential, then the direction of the electric field intensity taken should be positive. If it is directed from high to low potential, then the direction should be negative.