# How does electric field relate to voltage?

May 14, 2015

Electric field strength($E$) at a particular point in an Electric field(being generated by a charge $Q$), is defined as:

the Force($F$) that acts on a unit positive change($q$) placed at that point.

That is, $E = \frac{F}{q} = \frac{1}{4 \pi \epsilon} \frac{Q q}{r} ^ 2 \times \frac{1}{q} = \frac{1}{4 \pi \epsilon} \frac{Q}{r} ^ 2$

The magnitude of the Electric field($E$) can also be viewed as being the negative of the Potential gradient$\left(\frac{\mathrm{dV}}{\mathrm{dr}}\right)$

So $\text{ , "E=-(dV)/(dr) color(red)" {This is the relation i think}}$

Because $N {C}^{-} 1 \equiv V {m}^{-} 1$

where $V$ in $\frac{\mathrm{dV}}{\mathrm{dr}}$ is the Electric potential at the point under consideration,

And Electric Potential($V$) at a point in the E-field is the work done in bring a unit charge form infinity to that point

$\implies V = {\int}_{\infty}^{{r}_{1}} \frac{1}{4 \pi \epsilon} \frac{Q q}{r} ^ 2 \times \frac{1}{q} \equiv {\int}_{\infty}^{{r}_{1}} E \mathrm{dr}$

$\implies V = - \frac{1}{4 \pi \epsilon} \frac{Q}{{r}_{1}}$

So, it is logical that if, the $V = - \int E$, then the derivative $\frac{\mathrm{dV}}{\mathrm{dr}} = - E$