Electric potential — point charge, dipole, conducting sphere formulas

Question

What are the formulas for electric potential due to a point charge, an electric dipole, and a conducting sphere? How do we choose the right formula for a given charge configuration?

(CBSE 12, JEE Main, NEET — electric potential questions appear in every electrostatics paper, often combined with potential energy calculations)

Solution — Step by Step

V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r}

where $r$ = distance from the charge. Potential is a scalar — no direction, just sign.

Positive charge creates positive potential (decreases with distance)
Negative charge creates negative potential (increases toward zero with distance)
At infinity: $V = 0$ (reference point)

For multiple point charges: $V_{total} = V_1 + V_2 + V_3 + ...$ (algebraic sum, not vector sum — this is the advantage of working with potential instead of electric field).

V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p\cos\theta}{r^2}

where $p = qd$ = dipole moment, $\theta$ = angle between the dipole axis and the line joining the centre to the point.

Special cases:

On the axial line ( $\theta = 0$ ): $V = \frac{p}{4\pi\varepsilon_0 r^2}$ (maximum positive)
On the equatorial line ( $\theta = 90°$ ): $V = 0$ (the contributions from $+q$ and $-q$ cancel exactly)

Note the $1/r^2$ dependence — potential falls faster for a dipole than for a point charge ( $1/r$ ).

A conducting sphere of radius $R$ carrying charge $Q$ :

Outside the sphere ( $r > R$ ):

V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{r}

(Same as a point charge at the centre)

On the surface ( $r = R$ ):

V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{R}

Inside the sphere ( $r < R$ ):

V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{R} = \text{same as surface}

The potential inside a conductor is constant and equal to the surface potential. This is because $E = 0$ inside a conductor, and since $E = -dV/dr$ , $V$ must be constant.

flowchart TD
    A["Electric Potential Problem"] --> B{"What is the source?"}
    B -->|"Point charge"| C["V = kq/r<br/>Falls as 1/r"]
    B -->|"Dipole"| D["V = kp cosθ/r²<br/>Falls as 1/r²<br/>Zero on equatorial line"]
    B -->|"Conducting sphere"| E{"Where is the point?"}
    E -->|"Outside (r > R)"| F["V = kQ/r (like point charge)"]
    E -->|"On surface (r = R)"| G["V = kQ/R"]
    E -->|"Inside (r < R)"| H["V = kQ/R (constant)"]

Why This Works

Electric potential at a point represents the work done per unit positive charge in bringing a test charge from infinity to that point. For a point charge, this work depends only on the final distance $r$ (central force, path-independent). For a dipole, the two charges create partially cancelling potentials, giving the $\cos\theta$ angular dependence and faster $1/r^2$ decay.

Inside a conductor, any net electric field would cause charges to move (since they are free). In equilibrium, the field must be zero, which forces the potential to be uniform throughout the interior.

Common Mistake

Students confuse electric field and electric potential inside a conducting sphere. Inside: $E = 0$ but $V \neq 0$ . The potential is constant (equal to surface value), not zero. Zero field does not mean zero potential — it means the potential is not changing. A common NEET question: “What is the potential at the centre of a charged conducting sphere?” Answer: $kQ/R$ (not zero).

Potential is a scalar, so use algebraic addition for multiple charges. This is much simpler than adding electric fields (vectors). When a problem asks for the point where potential is zero, set up $V_1 + V_2 = 0$ and solve — much easier than finding where the field is zero.

Question

Solution — Step by Step

Why This Works

Common Mistake

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