Equipotential Surface and Its Properties: A surface that has a constant value of potential throughout is known as an equipotential surface. The potential difference between two points on an equipotential surface is zero. Equipotential surfaces are a useful way to represent the potential distribution in an electric field graphically. These surfaces can be represented in two dimensions using lines to help us quantitatively visualise the electric potential in the region. Let us read further to determine the properties of equipotential surfaces.

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What are Equipotential Surfaces?

The word “Equipotential” is a combination of “Equal” and “Potential”. An equipotential surface is a surface that has the same value of potential throughout. It can be defined as the locus of all points in the space that have the same value of potential. We can associate equipotential surfaces across a region having an electric field.

Since any surface having the same electric potential at every point is called an equipotential surface. Therefore the work done to move a charge from one point to another over an equipotential surface is zero. Equipotential points are all the points present in the space around an electric field with the same magnitude of electric potential. If a curve or a line connects these points, it is referred to as an equipotential line, and when these points lie on a specific surface, such a surface is called an equipotential surface. Moreover, if all the equipotential points are distributed uniformly across a volume or three-dimensional space, it is referred to as equipotential volume.

Work Done in Moving a Charge

Work is required to move a charge from one point to another in a given region. When the given region has equipotential all over it thus, the potential energy is constant throughout an equipotential surface. Thus, the work required to move a charge between two points in an equipotential surface equals zero.

We know that the work done to move a charge from one point to another is equal to the product of charge and the change in potential between the two points. Thus, is a point charge \(q\) is moved from a point \(A\) to point \(B\) such that potential at \(A\) is \({V_A}\) and potential at \(B\) is \({V_B}\) across an equipotential surface. Then the work done can be given as:

\(W = q\left( {{V_B} – {V_A}} \right)\)

Since the surface is equipotential, \({{V_B} = {V_A}}\)

Thus, \(W = q.0 = 0\)

Properties of Equipotential Surfaces

1. The direction of the electric field is always perpendicular to an equipotential surface.
2. Two equipotential surfaces can never intersect each other.
3. The equipotential surfaces are in the shape of concentric spherical shells around a point charge.
4. The equipotential surfaces are the planes that are normal to the x-axis in a region around a uniform electric field.
5. The potential Inside a hollow charged spherical conductor is constant. Thus, a hollow conductor can be treated as an equipotential volume. Thus, no work is required to move a charge from the centre to the surface or across the sphere of such a conductor.
6. The direction of the equipotential surface is from the region of higher potential to the region of lower potential.
7. The equipotential surfaces around an isolated point charge are in the form of spheres.
8. The concentric spheres around a point charge individually represent different equipotential surfaces.
9. Any plane normal to the direction of a uniform electric field is an equipotential surface.
10. We can identify strong or weak fields by the spacing in between the regions of 1equipotential surfaces, i.e.
    \(E = \, – \frac{{dV}}{{dr}}\)
    \( \Rightarrow \,E \propto  – \frac{1}{{dr}}\)

Electric Field and Equipotential Surface

We know that at every point on an equipotential surface, electric field lines are perpendicular to it. It is because of the fact that the potential gradient in a direction parallel to an equipotential surface is zero; thus, \(E = \, – \frac{{dV}}{{dr}} = 0\)

The component of the electric field parallel to the equipotential surface is zero. We can also understand it as: If the direction of the electric field were not normal to the equipotential surface, then it will have a non-zero component along its surface. This means that work will be required to move a unit test charge against the direction of the component of the electric field. But it contradicts the fact that no work is required to move a test charge across the equipotential surface. Thus, the electric field should be normal to the equipotential surface at all points.

Equipotential Surfaces Around Different Charge Configurations

1. For a point charge:

The formula for the electric potential of a point charge, \(V = \frac{{kq}}{r}\)

 Where \(r\) is the radius of the equipotential surface thus, the equipotential lines are circles, and in three dimensions equipotential surface is a sphere centred about the point charge.

2. For an electric dipole

An electric dipole consists of two charges of equal magnitude but opposite polarity. It is at the axis between the two dipoles, perpendicular to the plane where the electric potential due to the dipole is zero. The electric potential of an electric dipole is symmetrical at the centre of the dipole.

3. For a parallel plate capacitor

Within parallel conducting plates, like those of a capacitor, the electric field is uniform and perpendicular to the plates of the capacitor. Thus the equipotential lines will be parallel to the plates of the capacitor.

Problems Based on the Equipotential Surface

Q.1. Consider an electron of mass \(m\) and charge \(e\) released from rest into a uniform electric field of magnitude \({10^6}\frac{N}{C}\).  Compute its acceleration. Find the time taken by an electron to attain a speed of \(0.1c\), where \(c\) is the velocity of light. Take \(m = 9.1 \times {10^{ – 31}}{\rm{kg}},\,e = 1.6 \times {10^{ – 19}}{\rm{C}}\) and \(c = 3 \times {10^8}\,{\rm{m/s}}\).
Solution: Force on electron, \(F = eF = 1.6 \times {10^{ – 19}} \times {10^6} = 1.6 \times {10^{ – 13}}{\rm{N}}\)
Acceleration of the electron: \(a = \frac{F}{m} = \frac{{1.6 \times {{10}^{ – 13}}{\rm{N}}}}{{9.1 \times {{10}^{ – 31}}{\rm{Kg}}}}\)
Thus, \(a = 1.8 \times {10^{17}}\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}\)
It is given that the initial velocity of the electron, \(u = 0\)
After a time, ‘\(t\)’, the final velocity, \(v = 0.1c\)
Using the equation of motion,
\(v = u + at\)
\(t = \frac{v}{a} = \frac{{0.1c}}{{1.8 \times {{10}^{17}}}} = \frac{{0.1 \times 3 \times {{10}^8}}}{{1.8 \times {{10}^{17}}}}\)
\(t = 1.7 \times {10^{ – 10}}{\rm{s}}\)

Q.2. A charged particle having a charge \(q = 1.4\,{\rm{mC}}\) moves a distance of \(1.4\,{\rm{m}}\) along an equipotential surface of \(10\,{\rm{V}}\). Calculate the work done by the field throughout this motion.
Solution: The expression gives the work done by the field, \(W =\,  – q.\Delta V\)
For an equipotential surface, \(\Delta V = 0\)
Thus, the work done, \(W =\,  – q.0 = 0\)
work done is zero.

Q.3. A positively charged particle having a charge \(q = 1.0{\rm{C}}\) accelerates through a uniform electric field of \(10\,{\rm{V/m}}\). If the charged particle starts from rest on an equipotential plane of \(5\,{\rm{V}}\). The particle moves on an equipotential plane of \(V = 1\,{\rm{V}}\) after \(t = 0.0002{\rm{s}}\). Calculate the distance travelled by the particle.
Solution: Charge on the particle, \(q = 1.0\,{\rm{C}}\)
Electric field, \(E = 10\,{\rm{V/m}}\)
Let the distance travelled by change, \(d\)
Work done in moving a positively charged particle in an equipotential surface is given by, \(W =  \,- q.\Delta V\)
Substituting the values given in the question,
\(W = \, – \left( {1.0{\rm{C}}} \right)\left( {1V = 5{\rm{V}}} \right) = 4{\rm{J}}\)
Work done in moving a charge in an electric field, \(W = qEd\)
\(4 = \left( {1.0} \right)\left( {10} \right)d\)
\(d = 0.4\,{\rm{m}}\)

Summary

An equipotential surface is a surface that has the same value of potential throughout. It can be defined as the locus of all points in the space that have the same potential value. The work required to move a charge between two points in an equipotential surface equals zero. The direction of the electric field is always perpendicular to an equipotential surface; thus, \(E =\,  – \frac{{dV}}{{dr}} = 0\), and two equipotential surfaces can never intersect each other. We can identify strong or weak fields by the spacing in between the regions of equipotential surfaces.

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Frequently Asked Questions

Q.1. What is an equipotential surface?
Ans: An equipotential surface is a surface that has the same value of potential throughout.

Q.2. What is the word required to move a charge on an equipotential surface?
Ans: The work required to move a charge on an equipotential surface is zero.

Q3. Draw the equipotential surface around an electric dipole.
Ans: The equipotential surface can be represented as:

Q.4. Write two properties of equipotential surfaces.
Ans: Properties of equipotential surfaces are: 
1. Two equipotential surfaces can not intersect.
2. The direction of the electric field is always perpendicular to the direction of the equipotential surface.

Q.5. Can there be a non-zero component of the electric field along an equipotential surface?
Ans: No, there can not be a non-zero component of the electric field along an equipotential surface.

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