Electric Potential (V) Due To A Uniformly Charged Circular Disc - Field Theory.

- Consider a circular disc of radius ‘a’ which carries a uniform surface charge density ρs , C /m2.

- Say the disk lies on x - y plane (or z = 0 plane) with its axis along the z axis as shown in the figure.

- We need to find out electric potential (V) due to a circular disk at a point P (0, 0, h) on the z axis (z > 0).

- Electric potential (V) at a point due to any surface charge (ρs) is given as:

- In this case,

ds = ρ dρ dφ

(Since it’s a disc, the varying terms are radius ρ and angle φ)

R = (ρ2 + h2)1/2

- Hence electric potential (V) is given as:

- On solving further the equation becomes

- As a → 0, electric potential (V) also tends to zero i.e. V → 0.

- Hence the electric potential at point (0, 0, h) is given as:


- Gauss's Law - Theory.

- Gauss's Law - Application To a Point charge.

- Gauss's Law - Application To An Infinite Line Charge.

- Gauss's Law - Application To An Infinite Sheet Charge.

- Gauss's Law - Application To a Uniformly Charged Sphere.

- Numericals / Solved Examples - Gauss's Law.

- Scalar Electric Potential / Electrostatic Potential (V).

- Relationship Between Electric Field Intensity (E) and Electrostatic Potential (V).

- Electric Potential Due To a Circular Disk.

- Electric Dipole.

- Numericals / Solved Examples - Electric Potential and Electric Dipole.

- Energy Density In Electrostatic Field / Work Done To Assemble Charges.

- Numericals / Solved Examples - Electrostatic Energy and Energy Density.

- Numericals / Solved Examples - Gauss's law...

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  1. I'm looking for the answer to this when the point in question is not along the axial line. It's out there for a ring, but not for a disk.

  2. I was just wondering. Since you used a definite integral, why do you have C? Doesn't it cancel out when you integrate it from 0 to a?


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