Question

There is a uniform electrostatic field in a region. The potential at various points on a small sphere central at P, in the region, is found to vary between the limits $589.0{\text{ v to 589}}{\text{.8 v}}{\text{.}}$ What is the potential at a point on the sphere whose radius vector makes an angle of $60^\circ$ with the direction of the field?
(A) $589.2\,v$
(B) $589.5\,v$
(C) $589.6\,v$
(D) $589.4\,v$

Answer 1

To solve this problem we should know the concepts of electric potential, electric potential difference, relation between the electric field and electric potential and about the potential gradient.

Complete step by step answer
Potential energy per unit charge is called an electric potential.
The amount of work done in bringing the point charge from infinity to the point is called the electric potential at a point. Work is done against the electric field.
The amount of work done in bringing the point charge from one point to another point is called the electric potential difference between two points. Work is done against the electric field.
The electric potential at an infinite distance is mostly zero.
The electric potential V is a scalar quantity and has no direction, whereas the electric field E is a vector and has direction
The electric potential difference is given by
$\Delta V = Edx\,\cos \theta$
Where,
$\Delta V$ is the change in potential
E is the electric field
dx is the change in distance
$\theta$ is the angle between the point and the direction of the electric field.
Relation between electric field and potential:
Let distance between A and B be dx. Work done in bringing unit positive charge from A to B is
$\Rightarrow dV{\text{ }} = {\text{ }}E.dx$
$\Rightarrow E = - \dfrac{{dv}}{{dx}}$
Where,
E is the electric field
dv is the change in potential
dx is the change in distance
The change of potential with distance is known as potential gradient, hence the electric field is equal to the negative gradient of potential.
The negative sign in the formula represents that the potential decreases in the direction of the electric field. The unit of electric intensity can also be expressed as $V{m^{ - 1}}$
Given,
The angle between the point and the direction of electric field is $60^\circ$
The potential at various points on sphere is between the limits $589.0{\text{ v to 589}}{\text{.8 v}}{\text{.}}$
The change in the potential is given by
$\Rightarrow dV = 589.8 - {\text{589}}{\text{.0}}$
$\Rightarrow dV = 0.8\,V{m^{ - 1}}{\text{ }} \to {\text{1}}$
We have seen that, the potential gradient is
$\Rightarrow dV{\text{ }} = {\text{ }}E.dx{\text{ }} \to {\text{2}}$
So, from 1 and 2
$\Rightarrow E.dx = 0.8V{m^{ - 1}}$
We know that, the electric potential difference is given by
$\Rightarrow \Delta V = Edx\,\cos \theta$
Substitute the known values
$\Rightarrow \Delta V = 0.8 \times \cos 60^\circ$
$\Rightarrow \Delta V = 0.8 \times 0.5$
$\Rightarrow \Delta V = 0.4V{m^{ - 1}}$
The potential difference is $0.4V{m^{ - 1}}$
The electric potential at the point is given by the sum minimum potential on the sphere and the potential difference.
$\Rightarrow 589.0 + 0.4$
$\Rightarrow 589.4V{m^{ - 1}}$

Hence the correct answer is option (D) $589.4V{m^{ - 1}}$

Note: We should know the difference that the electric potential and electric potential difference are different from each other. The potential difference is potential between two point charges and electric potential is the potential due to the point charge or charges.