Symmetry-Based Electric Potential of Continuous Bodies: Formulating and establishing relations between the electric field and electrostatic potential on the axis of continuous charged bodies like semi-circular rings, discs, or uniformly charged thin rods

Concept

Electric potential due to a continuous charge distribution is obtained by adding the potential contribution of every small charge element.

dV = (1/4πε₀) dQ/r

Therefore, for a continuous distribution:

V = (1/4πε₀) ∫ dQ/r

Important: Electric potential is a scalar quantity. Therefore, potential contributions are added algebraically, while electric field is a vector and must be added vectorially.

E = -dV/dx

For an axis of symmetry, the field is usually along the axis. Hence:

E = -dV/dx

1. Uniformly Charged Ring

Consider a ring of radius R carrying total charge Q. Let P be a point on its axis at distance x from the centre.

Potential on the Axis

Every small charge element dq of the ring is at the same distance from P:

r = √(R² + x²)

Therefore:

dV = (1/4πε₀) dq/√(R²+x²)

Since the denominator is same for all charge elements:

V = (1/4πε₀) Q/√(R²+x²)

Potential on the axis:

V(x) = kQ/√(R²+x²)

Electric Field on the Axis

Using:

E = -dV/dx

We get:

E = kQx/(R²+x²)^3/2

Electric field on the axis:

E(x) = kQx/(R²+x²)^3/2

2. Uniformly Charged Semi-Circular Ring

Consider a uniformly charged semicircular ring of radius R and total charge Q. A point P is located on its symmetry axis at distance x from the centre.

Potential

Every charge element is at the same distance:

r = √(R²+x²)

Hence:

V = kQ/√(R²+x²)

The potential expression has the same form as that of a complete ring if Q represents the total charge actually present on the semicircular ring.

Electric Field

The field can be obtained from:

E = -dV/dx

Therefore, the axial component is:

E_x = kQx/(R²+x²)^3/2

However, because a semicircular ring is not completely symmetric like a full ring, components perpendicular to the axis may not cancel.

Important: For a complete ring, transverse components cancel by symmetry. For a semicircular ring, the geometry must be considered carefully before using the simple axial result.

3. Uniformly Charged Disc

Consider a uniformly charged disc of radius R with surface charge density σ. Let P be a point on the axis at distance x from its centre.

Divide the disc into thin rings of radius r and thickness dr.

dq = σ(2πr dr)

The distance from the ring to P is:

√(x²+r²)

Therefore:

dV = k(2πσr dr)/√(x²+r²)

Integrating from r = 0 to R:

V = 2πkσ ∫₀ᴿ r dr/√(x²+r²)

The result is:

V = (σ/2ε₀) [√(R²+x²) - x]

for x > 0.

Electric Field

Using:

E = -dV/dx

we obtain:

E = (σ/2ε₀) [1 - x/√(R²+x²)]

This is the electric field on the axis of a uniformly charged disc.

4. Uniformly Charged Thin Rod

Consider a uniformly charged rod of length L and linear charge density λ. Let P be a point on the perpendicular axis at distance x from the centre.

For a small element dx:

dq = λ dx

The potential is obtained using:

V = k ∫ dq/r

For a point on the perpendicular bisector of the rod:

r = √(x²+y²)

Thus:

V = kλ ∫_-L/2^L/2 dy/√(x²+y²)

The result is:

V = 2kλ ln [ (L/2 + √(x²+L²/4))/x ]

The electric field can then be obtained from the potential using the appropriate spatial derivative:

E = -∇V

Relation Between Electric Field and Potential

E = -dV/dx

The negative sign means that electric field points in the direction of decreasing potential.

Body	Potential on Axis	Field from Potential
Ring	kQ/√(R²+x²)	-dV/dx
Disc	σ[√(R²+x²)-x]/2ε₀	-dV/dx
Continuous Rod	k∫dq/r	-dV/dx

Section A — 1 Mark
10 MCQs

Q1. Electric potential is a: 1 Mark

A) Scalar quantity

B) Vector quantity

C) Tensor quantity

D) Dimensionless quantity

Correct Answer: A

Electric potential has magnitude only, so it is a scalar quantity.

Q2. The relation between electric field and potential along x-axis is: 1 Mark

A) E = dV/dx

B) E = -dV/dx

C) E = Vx

D) E = V/x²

Correct Answer: B

E = -dV/dx

Q3. Potential due to a point charge Q at distance r is: 1 Mark

A) kQ/r

B) kQ/r²

C) kQr

D) kQ²/r

Correct Answer: A

V = kQ/r

Q4. Every point of a uniformly charged ring is at the same distance from an axis point located on its axis. 1 Mark

A) True

B) False

C) Only at centre

D) Never

Correct Answer: A

Q5. The potential at the centre of a charged ring is: 1 Mark

A) kQ/R

B) kQ/R²

C) Zero

D) kQR

Correct Answer: A

At x = 0:

V = kQ/R

Q6. At the centre of a uniformly charged ring, electric field is: 1 Mark

A) Maximum

B) Zero

C) Infinite

D) kQ/R²

Correct Answer: B

Due to symmetry, fields from opposite elements cancel.

Q7. Electric field points towards: 1 Mark

A) Increasing potential

B) Decreasing potential

C) Constant potential only

D) Infinite potential

Correct Answer: B

E = -dV/dx

The negative sign indicates decreasing potential.

Q8. The SI unit of electric potential is: 1 Mark

A) Newton

B) Joule

C) Volt

D) Coulomb

Correct Answer: C

Q9. The potential due to a continuous charge distribution is found using: 1 Mark

A) Integration

B) Differentiation only

C) Multiplication only

D) Division only

Correct Answer: A

V = k∫dq/r

Q10. For a disc, a useful element for calculating potential is: 1 Mark

A) Small square

B) Thin ring

C) Point only

D) Sphere

Correct Answer: B

A disc can conveniently be divided into concentric thin rings.

Section B — 2 Marks
10 Conceptual + Numerical Questions

Q11. Find the potential at the centre of a ring of radius R carrying charge Q. 2 Marks

Answer:

V = kQ/R

At the centre every charge element is at distance R:

V = k∫dq/R = kQ/R

Q12. Write the potential at an axial point x from a charged ring. 2 Marks

V = kQ/√(R²+x²)

Q13. Find the electric field on the axis of a charged ring using V = kQ/√(R²+x²). 2 Marks

E = -dV/dx

Therefore:

E = kQx/(R²+x²)^3/2

Q14. What is the potential at the centre of a uniformly charged disc of radius R and surface charge density σ? 2 Marks

V = σR/(2ε₀)

Put x = 0 in:

V = σ[√(R²+x²)-x]/2ε₀

Q15. What is the electric field just above the centre of a very large charged disc? 2 Marks

E = σ/(2ε₀)

For R → ∞:

E = σ/(2ε₀)

Q16. Why is potential easier to calculate than electric field for many continuous charge distributions? 2 Marks

Potential is a scalar, so contributions can be added algebraically. Electric field is a vector and requires vector addition.

Q17. What is the potential at a point on the axis of a ring when x becomes very large? 2 Marks

V ≈ kQ/x

For x >> R:

√(R²+x²) ≈ x

Therefore:

V ≈ kQ/x

Q18. What happens to the electric field of a ring at x = 0? 2 Marks

E = 0

E = kQx/(R²+x²)^3/2

At x = 0, E = 0.

Q19. Why is the potential of a charged ring non-zero at its centre although the electric field is zero? 2 Marks

Electric field depends on vector cancellation, whereas potential is a scalar. Thus the fields cancel at the centre, but the potentials add.

Q20. Write the potential-energy relation for a charge q placed at potential V. 2 Marks

U = qV

Section C — 3 Marks
10 CBSE + Competitive Questions

Q21. Derive the electric potential on the axis of a uniformly charged ring. 3 Marks

For an element dq:

dV = k dq/√(R²+x²)

Since distance is the same for every element:

V = k/√(R²+x²) ∫dq

Since:

∫dq = Q

Therefore:

V = kQ/√(R²+x²)

Q22. Derive the electric field on the axis of a charged ring using the potential expression. 3 Marks

Given:

V = kQ(R²+x²)^-1/2

Using:

E = -dV/dx

Differentiate:

E = kQx(R²+x²)^-3/2

Hence:

E = kQx/(R²+x²)^3/2

Q23. Derive the potential on the axis of a uniformly charged disc. 3 Marks

Take a thin ring:

dq = σ2πrdr

Distance:

l = √(x²+r²)

Therefore:

dV = k(2πσrdr)/√(x²+r²)

Integrating from 0 to R:

V = σ[√(R²+x²)-x]/(2ε₀)

Q24. Obtain the electric field on the axis of a uniformly charged disc. 3 Marks

Start with:

V = σ[√(R²+x²)-x]/(2ε₀)

Use:

E = -dV/dx

Therefore:

E = σ/(2ε₀) [1 - x/√(R²+x²)]

Q25. Show that the electric field of a charged ring approaches the field of a point charge for x >> R. 3 Marks

For x >> R:

(R²+x²)^3/2 ≈ x³

Hence:

E ≈ kQx/x³

Therefore:

E ≈ kQ/x²

This is the electric field of a point charge Q.

Q26. Show that the potential of a charged ring approaches kQ/x far from the ring. 3 Marks

Given:

V = kQ/√(R²+x²)

For x >> R:

√(R²+x²) ≈ x

Hence:

V ≈ kQ/x

Q27. Find the potential at x = R on the axis of a charged ring. 3 Marks

Use:

V = kQ/√(R²+x²)

Put x = R:

V = kQ/√(2R²)

Therefore:

V = kQ/(R√2)

Q28. Find the ratio of the potential at the centre and at x = R for a charged ring. 3 Marks

At centre:

V₀ = kQ/R

At x = R:

V_R = kQ/(R√2)

Therefore:

V₀/V_R = √2

Answer:

√2 : 1

Q29. Explain why the electric field is zero at the centre of a uniformly charged complete ring. 3 Marks

Consider any small charge element dq on the ring. There is an identical element on the opposite side. Their electric fields at the centre have:

• Equal magnitudes
• Opposite directions

Therefore they cancel pairwise.

E_centre = 0

However, their potentials add because potential is scalar.

Q30. Explain the general method for finding electric field from potential for a continuous charged body. 3 Marks

Step 1: Choose a suitable coordinate along the symmetry axis.

Step 2: Calculate the potential:

V = k∫dq/r

Step 3: Simplify the integral using symmetry.

Step 4: Differentiate the potential:

E = -dV/dx

Step 5: Use the negative sign to determine the direction of the electric field.

Key Relation:

E = -dV/dx

Quick Revision Formula Sheet

Continuous Charge Distribution

V = k∫dq/r

Electric Field-Potential Relation

E = -dV/dx

Charged Ring

V = kQ/√(R²+x²)

E = kQx/(R²+x²)^3/2

Charged Disc

V = σ[√(R²+x²)-x]/(2ε₀)

E = σ/(2ε₀) [1-x/√(R²+x²)]

At Large Distance

V ≈ kQ/x

E ≈ kQ/x²

Exam Tip: For continuous charge distributions, calculate potential first whenever the geometry makes the distance from each element simple. Then obtain the electric field using E = -dV/dx.

Symmetry-Based Electric Potential of Continuous Bodies
CBSE + Competitive Physics Practice | 30 Questions with Answers & Solutions