JEE Main & Advanced Physics Electrostatics & Capacitance Electric Field

Electric Field

Category : JEE Main & Advanced

A positive charge or a negative charge is said to create its field around itself. Thus space around a charge in which another charged particle experiences a force is said to have electrical field in it.

(1) Electric field intensity \[(\vec{E})\]: The electric field intensity at any point is defined as the force experienced by a unit positive charge placed at that point. \[\vec{E}=  \frac{{\vec{F}}}{{{q}_{0}}}\]

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 \[\vec{E}=\underset{{{q}_{0}}\to 0}{\mathop{\text{Lim}}}\,\,\,\,\frac{{\vec{F}}}{{{q}_{\mathbf{0}}}}\]

(2) Unit and Dimensional formula

It's S.I. unit \[\frac{Newton}{coulomb}=\frac{volt}{meter}=\frac{Joule}{coulomb\times meter}\]

and C.G.S. unit - dyne/stat coulomb.

Dimension :\[[E]=[ML{{T}^{-3}}{{A}^{-1}}]\]

(3) Direction of electric field : Electric field (intensity) \[\vec{E}\] 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.

(4) Relation between electric force and electric field : In an electric field \[\vec{E}\] a charge (Q) experiences a force \[\overrightarrow{F}=Q\overrightarrow{E}\]. If charge is positive then force is directed in the direction of field while if charge is negative force acts on it in the opposite direction of field

(5) Super position of electric field (electric field at a point due to various charges) : The resultant electric field at any point is equal to the vector sum of electric fields at that point due to various charges i.e. \[\vec{E}={{\vec{E}}_{1}}+{{\vec{E}}_{2}}+{{\vec{E}}_{3}}+...\]

(6) Electric field due to continuous distribution of charge : A system of closely spaced electric charges forms a continuous charge distribution. To find the field of a continuous charge distribution, we divide the charge into infinitesimal charge elements. Each infinitesimal charge element is then considered, as a point charge and electric field \[\overrightarrow{dE}\] is determined due to this charge at given point. The Net field at the given point is the summation of fields of all the elements. i.e.,  \[\overrightarrow{E\,}=\int{\overrightarrow{dE}}\].  

Other Topics


You need to login to perform this action.
You will be redirected in 3 sec spinner