Tag: Potential due to a point charge

Derivation: Electrostatic Potential Due to a Point Charge

Step 1: Physical situation

Consider a point charge $Q$ placed at the origin O.
We want to find the electrostatic potential $V$ at a point P, which is at a distance $r$ from the charge.

By definition, electrostatic potential at a point is the work done per unit positive test charge in bringing it from infinity to that point, without acceleration.

Electric force on a unit test charge

For a point charge $Q$, electric field at distance $r^{\mathrm{\prime }}$ is

$$E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^{\mathrm{\prime }2}}$$

Since the test charge is unit positive charge, the force is

$$F=E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^{\mathrm{\prime }2}}$$

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Step 2: Small work done (why the minus sign appears)

Work done by external force for a small displacement $d{r}^{\mathrm{\prime }}$:

$$dW={\vec{F}}_{ext}\cdot d\vec{r}$$

The electric force pushes the charge away from Q, but we move it towards Q, so

$$dW=-Fd{r}^{\mathrm{\prime }}$$

$$dW=-\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^{\mathrm{\prime }2}}d{r}^{\mathrm{\prime }}$$

This minus sign is very important conceptually.

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Step 3: Setting up the definite integral

We bring the charge:

• from infinity → where potential is zero

• to distance $r$

So,

$$W={\int }_{\mathrm{\infty }}^{r}dW$$

$$W=-\frac{Q}{4\pi {\epsilon }_0}{\int }_{\mathrm{\infty }}^r\frac{d{r}^{\mathrm{\prime }}}{r^{\mathrm{\prime }2}}$$

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Step 4: Actual integration (expanded)

Recall basic calculus:

$$\int r^{\mathrm{\prime }-2}d{r}^{\mathrm{\prime }}=-r^{\mathrm{\prime }-1}$$

So,

$$W=-\frac{Q}{4\pi {\epsilon }_0}{[-\frac{1}{r^{\mathrm{\prime }}}]}_{\mathrm{\infty }}^r$$

Minus × minus becomes plus:

$$W=\frac{Q}{4\pi {\epsilon }_0}[\frac{1}{r}-\frac{1}{\mathrm{\infty }}]$$

Since $\frac{1}{\mathrm{\infty }}=0$,

$$\boxed{{\displaystyle W=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}}}$$

This is the total work done in bringing one unit charge from infinity to distance $r$

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2. Connecting Work Done and Potential (Key Concept)

Definition of Potential

$$\boxed{{\displaystyle V=\frac{W}{q}}}$$

Here:

• $W$ = work done

• $q$ = test charge

Since we used a unit charge ($q=1$),

$$V=W$$

So,

$$\boxed{{\displaystyle V(r)=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}}}$$

👉 Same mathematical expression, but different physical meaning.

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3. Why Work and Potential Have Different Units

This is where students usually mix things up.

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(a) Unit of Work Done

$$W=\text{Force}\times \text{distance}$$

$$\Rightarrow \text{unit of }W=\text{newton}\times \text{metre}$$

$$\boxed{{\displaystyle [W]=\text{joule (J)}}}$$

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(b) Unit of Potential

From definition,

$$V=\frac{W}{q}$$

$$\Rightarrow [V]=\frac{\text{joule}}{\text{coulomb}}$$

$$\boxed{{\displaystyle [V]=\text{volt (V)}}}$$