Question

Unit of electric field intensity is not:
(A) $\dfrac{V}{m}$
(B) $\dfrac{N}{C}$
(C) $\dfrac{{dyne}}{{statcoulomb}}$
(D) $\text{None of these}$

Answer 1

The electric field intensity at any point is defined as the force experienced by a unit positive charge placed at that point.

Complete solution:
1. A positive charge or a negative charge is said to create its field around itself. If a charge ${Q_1}$ exerts a force on charge ${Q_2}$ placed near it, it may be stated that since ${Q_2}$ is in the field of ${Q_1}$, it experiences some force, or it may also be said that since charge ${Q_1}$ is inside the field of ${Q_2}$, it experience some force. Thus space around a charge in which another charged particle experiences a force is said to have an electrical field in it.
2. The electric field intensity at any point is defined as the force experienced by a unit positive charge placed at that point.
$\overrightarrow E = \dfrac{{\overrightarrow F }}{{{q_0}}}$ Where ${q_0} \to 0$ so that presence of this charge may not affect the source charge Q and its electric field is not changed, therefore expression for electric field intensity can be better written as
$\overrightarrow E = \mathop {\lim }\limits_{{q_0} \to 0} \dfrac{{\overrightarrow F }}{{{q_0}}}$.
3. So, from these formulas the S.I unit of electric force intensity is
$\dfrac{F}{q} = \dfrac{{Newton}}{{Coulomb}} = \dfrac{N}{C} = \dfrac{{Volt}}{{meter}} = \dfrac{V}{m} = \dfrac{{Joule}}{{coulomb*meter}}$ and $C.G.S.$ $unit$- $\dfrac{dyne}{statcoulomb}$.

Hence from the given options (D) is correct.

Note: Electric field intensity is a vector quantity. Electric field due to a positive charge is always away from the charge and that due to a negative charge is always towards the charge. The stat coulomb is defined as if two stationary objects each carry a charge of 1 stat coulomb and are 1 cm apart, they will electrically repel each other with a force of 1 dyne.