Question
Derive relation between electric field $(E)$ and electric potential $(V).$
✓
Answer
$i.$ A pair of parallel plates is connected as shown in the figure. The electric field between them is uniform
$ii.$ A potential difference $V$ is applied between two parallel plates separated by a distance $‘d \ ’.$
$iii.$ The electric field between them is directed from plate $A$ to plate $B.$
$iv.$ A charge $+q$ placed between the plates experiences a force $F$ due to the electric field.
$v.$ If the charge is moved against the direction of field, i.e., towards the positive plate, some amount of work is done on it.
$vi.$ If the charge is moved $+q$ from the negative plate $B$ to the positive plate $A,$ then the work done against the field is $W = Fd;$ where $‘d \ ’$ is the separation between the plates.
$vii.$ The potential difference $V$ between the two plates is given by $W = V_q,$
but $W = Fd$
$\therefore Vq = Fd$
$\therefore \frac{F}{q}=\frac{V}{d}= E$
$\therefore$ Electric field can be defined as $E = V/d.$
$ii.$ A potential difference $V$ is applied between two parallel plates separated by a distance $‘d \ ’.$
$iii.$ The electric field between them is directed from plate $A$ to plate $B.$
$iv.$ A charge $+q$ placed between the plates experiences a force $F$ due to the electric field.
$v.$ If the charge is moved against the direction of field, i.e., towards the positive plate, some amount of work is done on it.
$vi.$ If the charge is moved $+q$ from the negative plate $B$ to the positive plate $A,$ then the work done against the field is $W = Fd;$ where $‘d \ ’$ is the separation between the plates.
$vii.$ The potential difference $V$ between the two plates is given by $W = V_q,$
but $W = Fd$
$\therefore Vq = Fd$
$\therefore \frac{F}{q}=\frac{V}{d}= E$
$\therefore$ Electric field can be defined as $E = V/d.$
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