Derive an expression for electrostatic potential due to a point charge.

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Q. Derive an expression for electrostatic potential due to a point charge.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2023Subjective· 3mImportance★★★★★

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Concept understanding — Electric Potential

Electric Potential

The Idea

When you lift a book onto a shelf you do work against gravity, and the book gains gravitational potential energy that depends on where it sits. Electric charges behave the same way in an electric field. Electric potential is the electrical analogue of "height" in a gravitational field: it tells you the potential energy a unit charge would have at a given point.

Formally, the electric potential at a point is the work done in bringing a unit positive charge from infinity to that point, slowly (without giving it kinetic energy), against the electric field.

V = \frac{W}{q_0}

Here $W$ is the work done to move a small test charge $q_0$ from infinity to the point. Because both $W$ and $q_0$ are scalars, electric potential is a scalar quantity — it has magnitude but no direction. Its SI unit is the volt:

1\ \text{V} = 1\ \text{J/C}

Potential Due to a Point Charge

For a single point charge $q$ , the potential at a distance $r$ from it is

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

Notice it falls off as $1/r$ , whereas the electric field of a point charge falls off as $1/r^2$ . The potential is taken as zero at infinity, our chosen reference. A positive charge makes the potential around it positive; a negative charge makes it negative.

Potential Difference

Usually we care about the potential difference between two points $A$ and $B$ :

V_A - V_B = \frac{W_{B\to A}}{q_0}

the work per unit charge needed to move a charge from $B$ to $A$ . This is the quantity a voltmeter reads and the "voltage" that drives current in a circuit.

Relation Between Field and Potential

Field and potential are two views of the same thing. The electric field is the negative rate of change of potential with distance:

E = -\frac{dV}{dr}

The minus sign says the field points from high potential toward low potential — a positive charge, left free, rolls "downhill" in potential. Where the potential changes steeply, the field is strong.

Superposition

Because potential is a scalar, the potential due to several charges is just the algebraic sum of their individual potentials — no vectors, no angles:

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \cdots\right)

This makes potential far easier to compute than the field, which needs vector addition. Once you have $V$ everywhere, you can get the field by differentiating.

Equipotential Surfaces

An equipotential surface is a surface on which the potential has the same value everywhere.

No work is done in moving a charge along an equipotential surface (since $\Delta V = 0$ ).
The electric field is always perpendicular to an equipotential surface. …

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The potential of a point charge $q$ at distance $r$ is obtained by integrating the electric field from infinity (zero potential) up to $r$ , giving $V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}$ .

1. Setup. Consider a point charge $q$ placed at the origin $O$ . We wish to find the electrostatic potential $V$ at a point $P$ located at distance $r$ from $O$ . By convention, potential is taken to be zero at infinity.

2. Electric field due to the point charge. At a distance $x$ from $O$ , the Coulomb field is

$E(x) = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{x^2}$

directed radially outward (for $q>0$ ).

3. Relation between field and potential. The potential difference is related to the field by

$V(r) - V(\infty) = -\int_{\infty}^{r} E(x)\,dx$

Since $V(\infty)=0$ ,

$V(r) = -\int_{\infty}^{r} \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{x^2}\,dx$

4. Evaluating the integral.

$V(r) = -\dfrac{q}{4\pi\varepsilon_0}\int_{\infty}^{r} x^{-2}\,dx = -\dfrac{q}{4\pi\varepsilon_0}\Big[-\dfrac{1}{x}\Big]_{\infty}^{r}$ …

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Q.Derive an expression for electrostatic potential due to a point charge.

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