Question

Derive $E = \dfrac{{ - dV}}{{dr}}$ , where the symbols have their usual meaning.

Answer 1

Hint : In order to this question, to explain the derivation of the given equation $E = \dfrac{{ - dV}}{{dr}}$ , we will first write the formula of velocity of the charged particle and then we will differentiate with respect to its radius. And then we will write the equation of Electric field, as we will see that we get our given equation.

Complete Step By Step Answer:
As we know that, for the charged particle, the formula of its velocity is-
$V = \dfrac{{K{Q^2}}}{r}$
where, $V$ is the velocity of charged particle
$Q$ is the charge on the particle
$r$ is the radius.
Now, If we differentiate the formula with respect to its radius, we get:
$\because \dfrac{{dv}}{{dr}} = \dfrac{{ - k{Q^2}}}{{{r^2}}} \\\ \Rightarrow - \dfrac{{dv}}{{dr}} = \dfrac{{k{Q^2}}}{{{r^2}}}$ ……..eq(ii)
As we know that: the formula of the Electric field:
$E = \dfrac{{K{Q^2}}}{{{r^2}}}$ ……..eq(ii)
Now, from eq(i) and eq(ii):-
we get-
$\therefore E = - \dfrac{{dV}}{{dr}}$ .

Note :
$E = - \dfrac{{dV}}{{dr}}$ is always true. Only when computing the size of the E field between closely spaced parallel plates does $E = \dfrac{V}{r}$ apply. The first equation's negative sign indicates that the Electric field is pointing in the opposite direction as the voltage drops. As a result, if voltage decreases in the direction of $r$ , the Electric field will point in the same direction.