Question

A 5C charge experiences the force of 2000 N when moved between two points on a field separated by the distance of 2 cm in a uniform electric field. The potential difference between the two points is:
A. 8 volts
B. 80 volts
C. 800 volts
D. 8000 volts

Answer 1

Hint: In this question we have been asked to calculate the potential difference across two points on a field separated by 2 cm distance. We know that potential across two points is given by Electric field over the distance between two points. We are given the distance, therefore, we will calculate the electric field using given information.
Formula Used:
\[F=qE\]
Where,
F is he force on charge in Newton
Q is the charge in coulombs
E is the net electric field.
\[V=E.d\]
Where,
V is the potential difference in volts
d is the distance between two points in meters

Complete answer:
We know the distance between the two points. Therefore, to calculate the electric field
 

We know,
\[F=qE\]
Therefore,
\[E=\dfrac{F}{q}\]
After substituting the given values
We get,
\[E=\dfrac{2000}{5}\]
Therefore,
\[E=400\]
We have been given distance is 2 cm which can be taken as 0.02m
Therefore, we know the potential across the point is given by
\[V=E.d\]
After substituting values
We get,
\[V=400\times 0.02\]
Therefore,
\[V=8\]volts

Therefore, the correct answer is option A.

Note:
We know that electric field is defined as electric force per unit charge. The direction of the field is usually taken as the direction of force exerted. It is radially outward from positive charge and radially inward for negative charge. The measure of energy transferred in between two points is known as the potential between two points. It is generated by the charge in the circuit.