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Relation Between Electric Field Intensity And Electric potential

Relation Between Electric Field Intensity And Electric potential :- From the definition, potential difference between two points A and B is given by :

$\displaystyle {{V}_{B}}-{{V}_{A}}=\frac{{{W}_{AB}}}{{{q}_{o}}}$ …..(1)

If the points A and B are very close ( $\displaystyle \overrightarrow{dr}$ ) to each others, then

$\displaystyle {{V}_{B}}-{{V}_{A}}=dV$

$\displaystyle \Rightarrow dV=\frac{{{W}_{AB}}}{{{q}_{o}}}=\frac{\left( \overrightarrow{{{F}_{ext}}} \right).\left( \overrightarrow{dr} \right)}{{{q}_{o}}}$

$\displaystyle \Rightarrow dV=\frac{\left( -{{q}_{o}}\overrightarrow{E} \right).\left( \overrightarrow{dr} \right)}{{{q}_{o}}}$

$\displaystyle dV=-\overrightarrow{E}.\overrightarrow{dr}$ …..(2)

Integrating equation (2) we get,

$\displaystyle \int{dV=-\int{\overrightarrow{E}.\overrightarrow{dr}}}$

$\displaystyle V=-\int\limits_{\infty }^{r}{\overrightarrow{E}.\overrightarrow{dr}}$ …..(3)

In equation (2), the the small displacement ( $\displaystyle \overrightarrow{dr}$ ) and electric field ( $\displaystyle \overrightarrow{E}$ ) can be written as,

$\displaystyle \overrightarrow{dr}=dx\widehat{i}+dy\widehat{j}+dz\widehat{k}$ …..(4)

$\displaystyle \overrightarrow{E}={{E}_{x}}\widehat{i}+{{E}_{y}}\widehat{j}+{{E}_{z}}\widehat{k}$ …..(5)

Using equations (4) and (5) in equation (2) :-

$\displaystyle dV=-\overrightarrow{E}.\overrightarrow{dr}=-\left( {{E}_{x}}\widehat{i}+{{E}_{y}}\widehat{j}+{{E}_{z}}\widehat{k} \right).\left( dx\widehat{i}+dy\widehat{j}+dz\widehat{k} \right)$

$\displaystyle dV=-\left( {{E}_{x}}dx+{{E}_{y}}dy+{{E}_{z}}dz \right)$ …..(6)

Now the total differential of is :-

$\displaystyle dV=\frac{\partial V}{\partial x}dx+\frac{\partial V}{\partial y}dy+\frac{\partial V}{\partial z}dz$ …..(7)

Comparing equations (6) and (7),

$\displaystyle {{E}_{x}}=-\frac{\partial V}{\partial x}\text{, }{{E}_{y}}=-\frac{\partial V}{\partial y},\text{ }{{E}_{z}}=-\frac{\partial V}{\partial z}$

Putting the values of E_x , E_y and E_z in equation (5),

$\displaystyle \overrightarrow{E}=-\frac{\partial V}{\partial x}\widehat{i}-\frac{\partial V}{\partial y}\widehat{j}-\frac{\partial V}{\partial z}\widehat{k}$

$\displaystyle \overrightarrow{E}=-\left( \frac{\partial }{\partial x}\widehat{i}+\frac{\partial }{\partial y}\widehat{j}+\frac{\partial }{\partial z}\widehat{z} \right)V$

$\displaystyle \Rightarrow \overrightarrow{E}=-\overrightarrow{\nabla }\text{ V}$ …..(8)

Equation (8) shows that the electric field is the negative gradient of the electric potential.

Here $\displaystyle \overrightarrow{\nabla }=del\text{ }operator$ . It is denoted by the nabla symbol is a vector differential operator used in vector calculus.

Relation Between Electric Field Intensity And Electric potential

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