Question

Deduce an expression for electric potential due to a charge at any point?

Answer 1

Electric Potential due to a Point Charge

According to definition electric potential at a point is equal to the amount of work done in bringing unit positive test charge from infinity to that point in the electric field without any acceleration.

Suppose a point charge $$ +q $$ is placed at point 0 and potential due to this charge at point $$ P $$ situated at distance r from the charge, is to be determined. According to definition of potential, to find out the potential at P, we have to estimate the work done in bringing $$ +1 \mathrm{C} $$ charge from infinity to point $$ \mathrm{P} $$ inside the electric field. For this purpose, let us consider another point $$ \mathrm{P}^{\prime} $$ situated at distance $$ \mathrm{x} $$ from $$ \mathrm{O} $$ beyond $$ \mathrm{P} $$ and $$ +\mathrm{q}_{0} $$

positive test charge is placed at $$ \mathrm{P} $$, the electrostatic force acting on $$ +\mathrm{q}_{0} $$, $$ F=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q q_{0}}{x^{2}} $$

Work done in giving a small displacement dx toward P,

$$\mathrm{dW}=\vec{F} \cdot \overrightarrow{d x}=\mathrm{F} \mathrm{dxcos} 180^{\circ}=\mathrm{F} \mathrm{dx}(-1)$$

or $$ \mathrm{dW}=-\mathrm{F} \mathrm{dx} $$

Therefore work done in bring $$ +\mathrm{q}_{0} $$ charge from infinity to $$ \mathrm{P} $$

$$W=\int d W=-\int_{\infty}^{r} F d x$$

or $$W=-\int_{\infty}^{r} \dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q q_{0}}{x^{2}} d x$$

$$\begin{array}{l}=-\dfrac{q q_{0}}{4 \pi \varepsilon_{0}} \int_{\infty}^{r} \dfrac{1}{x^{2}} d x \\=-\dfrac{q q_{0}}{4 \pi \varepsilon_{0}} \int_{\infty}^{r} x^{-2} d x \\=-\dfrac{q q_{0}}{4 \pi \varepsilon_{0}}\left[-\dfrac{1}{x}\right]_{\infty}^{r}\end{array}$$

$$\left[\text { Using formula } \int x^{n} d x=\dfrac{x^{n+1}}{n+1}\right]$$

but $$ \quad n \neq-1 $$

$$\begin{array}{c}=\dfrac{q q_{0}}{4 \pi \varepsilon_{0}}\left[\dfrac{1}{r}-\dfrac{1}{\infty}\right] \\{\left[\mathrm{U} \operatorname{sing}(x)_{x_{1}}^{x_{2}}=x_{2}-x_{1}\right]} \\\left.W=\dfrac{q q_{0}}{4 \pi \varepsilon_{0}} \dfrac{1}{r} \quad \text { (because } \dfrac{1}{\infty}=0\right)\end{array}$$

or $$ \quad \dfrac{W}{q_{0}}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{r} $$

or $$ \quad V=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{r} $$

Where $$ \vee \rightarrow $$ potential at $$ P $$

If q is positive then potential $$ \vee $$ will also be positive and if $$ q $$ is negative, then potential will also be negative."From equation ( $$ 1) $$

$$ \vee \propto \dfrac{1}{r} $$

Electric field due to point charge $$ E=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{r^{2}} \therefore E \propto \dfrac{1}{r^{2}} $$

If $$ \mathrm{V} $$ and $$ \mathrm{E} $$ both are plotted with distance r then the graphs will be as under figure $$ [3.4(\mathrm{a})] . $$

For negative charge $$ \vee-r $$ graph will be as under figure $$ [3.4(b)] $$