What is Equipotential Surface?

Equipotential Surface is a surface in an electric field where the potential at each point on the surface is the same.

An equipotential surface is a surface where the potential is the same at every point; no work is needed to move a charge from one point to another on the surface.

Table of Contents

• Work Done in Equipotential Surface
• Properties of Equipotential Surface
• Solved Examples

If the points in an electric field are all at the same electric potential, they are known as Equipotential Points. An Equipotential Line is formed when these points are connected by a line or a curve. If such points lie on a surface, it is called an Equipotential Surface. Additionally, if these points are distributed throughout a space or a volume, it is known as an Equipotential Volume.

Work Done in an Equipotential Surface

If a point charge is moved from point V_A to V_B, in an equipotential surface, then the work done in moving a charge between the two points is zero.

$W=q_0(V_A-V_B)$

As V_A - V_B is equal to zero, the total work done is W = 0.

Properties of Equipotential Surface

1. An equipotential surface is always perpendicular to an electric field.

2. Two equipotential surfaces cannot intersect.

3. For a point charge, the equipotential surfaces are concentric spherical shells.

4. Equipotential surfaces for a uniform electric field are planes normal to the x-axis.

5. The direction of the equipotential surface is from low potential to high potential.

6. A hollow charged spherical conductor is an equipotential volume, meaning that the potential is constant throughout and no work is required to move a charge from the centre to the surface.

7. For an isolated point charge, the equipotential surface is a sphere; that is, concentric spheres around the point charge are different equipotential surfaces.

In a uniform electric field, any plane perpendicular to the field direction is an equipotential surface.

**9. The spacing between equipotential surfaces allows us to differentiate between regions of a strong and weak field (i.e. $E=-\frac{dV}{dr}\Rightarrow E\propto \frac{1}{dr}$)

**

Also Read

• Electrostatics
• Electric Charge
• Gauss’s Law
• Coulomb’s Law
• Electric Field Intensity
• Electric Potential Energy
• Motion of a Charged Particle in an Electric Field

Problems on Equipotential Surface

Q.1: The work done by the field during the motion of the charged particle (q =1.4 mC) over a distance of 0.4 m along an equipotential surface of 10 V is 5.6 mJ.

Solution:

The expression that gives the work done by the field is given below.

-W = qΔV

Since ΔV = 0, the work done for equipotential surfaces is zero, W = 0.

Q.2: Determine the distance travelled by the particle after 0.0002 seconds, given that it started from rest on an equipotential plane of 50 V and is now on an equipotential plane of 10 V in a uniform electric field of 100 V/m and a positive charge of 1.0 C.

Solution:

The work done in moving a charge across an equipotential surface is given by

W = -qΔV

Substituting the values, we get:

$W=(-1.0,C)\times (10V-50V)=40J$

We know that the work done in moving a charge in an electric field is equal to the change in electric potential energy.

W = qEd

40 = 100%

d = 0.4\ m

Q.3: An electron of mass m = 9.1 × 10^–31 kg and charge e = 1.6 × 10^–19 coulomb is released from rest in a uniform electric field of 106 newton/coulomb. Compute its acceleration. Also, find the time taken by the electron to attain a speed of 0.1 c, where c is the velocity of light (c = 3 ×108 metre/sec).

Sol.

The force experienced by the electron is

F = E_e = 106 (1.6 × 10^-19)

1.6 × 10^-13 N

The acceleration of the electron is given by acceleration

\begin{array}{l}a = \frac{1.6\times {{10}^{-13}}}{9.1\times {{10}^{-31}}} \times \frac{m}{F}\end{array}

1.8 x 10¹⁷ m/sec²

The initial velocity is equal to zero.

Let t be the time taken by the electron to attain a final speed of 0.1c.

Now $$v = u + at \quad \text{or} \quad v = at$$

$$\therefore t = \frac{v}{a} = \frac{0.1\times (3\times {{10}^{8}})}{1.8\times {{10}^{17}}} = 1.67 \times {{10}^{-10}} \text{s}$$

1.7e10 sec.

Q.4: What is the horizontal velocity that must be imparted to a ball of mass m with a charge q, rotating in a vertical plane at the end of a string of length λ in a uniform electrostatic field whose lines of force are directed upwards, for the tension in the string in the lower position of the ball to be 15 times the weight of the ball?

Sol.

The situation depicted in Fig. is shown below.

At B, the sum of kinetic energy and potential energy is equal to the sum of kinetic energy and potential energy at A, as shown in the image below:

$$(K.E.)_A + (P.E.)_A = (K.E.)_B + (P.E.)_B ..(1)$$

Gain in K.E.

$$(K.E.)_A - (K.E.)_B = (1/2)m(v^2_2 - v^2_1) …(2)$$

$$(P.E.)_A - (P.E.)_B = Loss in P.E.$$

=Loss in Gravitational Potential Energy - Gain in Electrostatic Potential Energy

2mg - qE.2λ

$$(mg - qE)^2 \lambda \ldots$$(3)

From eq. (1),

$$(K.E.)_A – (K.E.)_B = (P.E.)_B – (P.E.)_A$$

Substituting the values, we get

$$(mg - qE) \times 2\lambda = \frac{1}{2}m(v_2^2 - v_1^2) \dots (4)$$

Centripetal Force at A = T² + qE - mg

$$\therefore T_2 + qE - mg = \frac{mv_2^2}{\lambda} $$

From eq. (4): $$mv_{22} = 2 \left(mg - qE \right) \frac{2\lambda}{mv_{12}}$$

From eq. (5), $$mv_{22} = \lambda \left(T_2 + qE - mg \right)$$

2(mg - qE) * 2λ - mv^2 = λ(T^2 + qE - mg)

\begin{array}{l}4qE - 4mg + \left( \frac{m}{\lambda} \right)v_{1}^{2} = T_{2} + qE - mg\end{array}

Putting T2 = 15 mg, we have

\begin{array}{l}4mg - 4qE + \left( \frac{m}{\lambda} \right) v_{1}^{2} = 15mg + qE - mg\end{array}

\begin{array}{l}\left( \frac{m}{λ} \right)v_{1}^{2} = 10mg + 5qE = 5 \cdot (2mg + qE)\end{array}

\begin{array}{l}v_1 = \sqrt{\frac{1}{m}.(2mg+qE),.,5}\end{array}

Q.5: Find the potential and electric field at the point x=0 due to an infinite set of charges, each equal to q, placed along the x-axis at x=1, x=2, x=4, x=8, and so on.

Sol. The electric potential V and electric field intensity E at a position x due to a point charge can be expressed as potential and field intensity respectively.

\begin{array}{l}V = \frac{q}{4\pi \varepsilon_{0} x} \text{and} E = \frac{q}{4\pi \varepsilon_{0} {x}^{2}}\end{array}

The principle of superposition holds good for electric interaction, which is a two body interaction.

\begin{array}{l}V = \frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \sum\limits_{i=1}^{\infty }\frac{q}{i} \right]\end{array}

\begin{array}{l}= \frac{q}{4\pi \varepsilon_0}\left(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots\right)\end{array}

I.e., this statement means that the response should be written in markdown.

\begin{array}{l}V = \frac{2q}{4\pi {{\varepsilon }_{0}}} = \frac{1}{2\pi {{\varepsilon }_{0}}}\end{array} (2q)

\begin{array}{l}E = \frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{q}{{{1}^{2}}}+\frac{q}{{{2}^{2}}}+\frac{q}{{{4}^{2}}}+,,… \right]\end{array}

\begin{array}{l}=\frac{q}{4\pi \varepsilon_0}\left[\sum_{n=0}^{\infty}\frac{1}{4^n}\right]\end{array}

I.e., this statement means that the example provided is an illustration of the concept.

\begin{array}{l}E = \frac{1}{4\pi {{\varepsilon }_{0}}} \left[ \frac{4}{3},q \right]\end{array}

Q.6: Find the work done in displacing a charge q = 1000 μC from a point (0, √(0.44) m) to (0, 1 m) on a thin rod of length λ = 1 m, lying along the x-axis with one end at the origin x = 0, where the uniformly distributed charge per unit length λ = Kx, and K = constant = 10^–9 cm^–2.

Sol.

The situation is shown in Fig. Consider a small element of length dx of the rod at a distance x from the origin. Then, the potential dVP at point P due to this element is given by

\begin{array}{l}K = \frac{dVP}{\frac{xdx}{r}} = \frac{4\pi {{\varepsilon }_{0}};dVP}{xdx} = 4\pi {{\varepsilon }_{0}}r;dVP\end{array}

From Figure,

r² = x² + y²

$$2\frac{\mathrm{d}r}{\mathrm{d}x} = 2x (y = \text{constant})$$

$$(\therefore dVP = \frac{K}{4\pi {{\varepsilon }_{0}}}dr)$$

On Integrating this expression, we get the potential at point P. Thus, it can be concluded that integration is essential for finding the potential at any given point.

\begin{array}{l}V_p = \frac{K}{4\pi {{\varepsilon }_{0}}};;\int_{y}^{\sqrt{({{λ }^{2}}+{{y}^{2}})}}{dr} \end{array}

$$(\frac{K}{4\pi \varepsilon_{0}} \left[\sqrt{\lambda^2 + y^2} - y\right])$$

$$(\therefore V_{A} = \frac{K}{4\pi {{\varepsilon }_{0}}}[\sqrt{1.44} - \sqrt{0.44}])$$

$$(\frac{0.5366 K}{4\pi \varepsilon_0})$$

= (9 × 10⁹) × 10^-9 × 0.5366 = 4.83 V

$$V_{B} = 0.4142\frac{K}{4\pi \varepsilon_{0}}$$

3.728 V = (9 × 10^9) × 10^-9× 0.4142

Net work done

$$=-1.1 \times 10^{-3} \text{ Joule}$$

Frequently Asked Questions about Equipotential Surfaces

An equipotential surface is a surface in an electric field where the potential is constant at every point.

A surface with an equipotential potential is one where all points on the surface have the same electric potential. This means that at every point on the equipotential surface, a charge will have the same potential energy.

What is the value of work done on an equipotential surface?

The value of work done on an equipotential surface is zero.

The work done in moving a charge between two points on an equipotential surface is zero.

Properties of an Equipotential Surface:

1. All points on an equipotential surface have the same electric potential.
2. An equipotential surface is always perpendicular to the electric field lines.
3. The electric field strength is zero at all points on an equipotential surface.
4. An equipotential surface has no electric field associated with it.

An equipotential surface has an electric field that is constantly perpendicular to it.

The intersection of two equipotential surfaces is impossible.

Equipotential surfaces for a point charge are concentric spherical shells.

Equipotential surfaces are planes normal to the x-axis in a homogeneous electric field.

The equipotential surface runs from high potential to low potential.

The potential inside a hollow charged spherical conductor is constant. Moving a charge from the centre to the surface requires no effort.