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Electric Potential Due To A Point Charge

Electric Potential Due To A Point Charge :- Let us consider a point charge placed at a fixed point O in space. We want to calculate the electric potential at a point a P located at a distance from this charge.

By definition, electric potential at a point is the work done by external agent in bringing a unit positive charge from infinity (where the potential is taken as zero) to that point, without any acceleration.

As electrostatic force is conservative in nature, work done is independent of the path followed. So we choose a convenient straight line path along the radial direction of charge from infinity to point P.

The small amount of work done in moving the a test charge +q₀ through a small displacement $\displaystyle \overrightarrow{dx}$ from point A to point B against the electric field $\displaystyle \overrightarrow{E}$ is given by :

$\displaystyle dW=\left( {{\overrightarrow{F}}_{ext}} \right).\left( \overrightarrow{dx} \right)$

$\displaystyle dW=\left( {{\overrightarrow{F}}_{ext}} \right).\left( \overrightarrow{dx} \right)={{F}_{ext}}\text{ }dx\text{ }cos0={{F}_{ext}}dx$

As we have to move the test charge +q₀ without any acceleration, so magnitude of external force must be equal to the magnitude of electrostatic force, i.e.,

$\displaystyle \left| {{\overrightarrow{F}}_{ext}} \right|=\left| {{\overrightarrow{F}}_{E}} \right|={{q}_{0}}E$

Hence can be written as

$\displaystyle dW={{F}_{ext}}dx={{q}_{0}}Edx$

Putting the value of electric filed at point A, $\displaystyle E=\frac{kQ}{{{x}^{2}}}$ , dW becomes :

$\displaystyle dW={{q}_{0}}Edx={{q}_{0}}\frac{kQ}{{{x}^{2}}}dx$

Now to calculate total work done in moving the test charge +q₀ from infinity to point P, we have to integrate within limits x = ∞ to x = r. But note that as we move from infinity to point P, we are actually moving in a direction in which x decreases, so we have to replace (dx) by (- dx) and then we get total work. Hence,

$\displaystyle W=\int\limits_{\infty }^{r}{\frac{k{{q}_{0}}Q}{{{x}^{2}}}\left( -dx \right)}=-k{{q}_{0}}Q\int\limits_{\infty }^{r}{{{x}^{-2}}}dx=-k{{q}_{0}}Q\left[ \frac{{{x}^{-1}}}{-1} \right]_{\infty }^{r}$

$\displaystyle \Rightarrow W=k{{q}_{0}}Q\left[ \frac{1}{x} \right]_{\infty }^{r}=k{{q}_{0}}Q\left[ \frac{1}{r}-\frac{1}{\infty } \right]=k{{q}_{0}}Q\left[ \frac{1}{r}-0 \right]$

$\displaystyle W=\frac{k{{q}_{0}}Q}{r}$

Now by definition, electric potential due to a point charge at point P,

$\displaystyle V=\frac{{{W}_{\infty P}}}{{{q}_{0}}}=\frac{W}{{{q}_{0}}}=\frac{k{{q}_{0}}Q}{r}\times \frac{1}{{{q}_{0}}}$

$\displaystyle \therefore V=\frac{kQ}{r}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{r}$ …..(1)

When Q is positive, electric potential V is positive and when Q is negative, electric potential V is negative. At r = ∞, V = Q/r = 0, i.e., electric potential due to a point charge is zero at infinity.

From equation (1) we can say that at equal distances from a point charge Q, value of V is same. Hence, electric potential due to a single point charge is spherically symmetric.

Below graph shows the variation of electric potential V with distance r ( $\displaystyle V\propto \frac{1}{r}$ ) and variation of electric field intensity E with distance r ( $\displaystyle E\propto \frac{1}{{{r}^{2}}}$ ) from point charge Q :

Electric Potential Due To A Point Charge

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