# Electric Field Due To Point Electric Charges

“Every charge in the universe exerts a force on every other charge in the universe” is a bold yet true statement of physics. One way to understand the ability of a charge to influence other charges anywhere in space is by imagining the influence of the charge as a field. Note that this ‘influence’ is simply the electrostatic force that a charge is able to exert over another. However when describing fields, we require a quantity (scalar or vector), that is independent of the charge it is acting on and only dependent on the influence and the spatial distribution.

So in a simple way we can define the electrostatic field considering the force exerted by a point charge on a unit charge. In other words we can define the electric field as the force per unit charge.

To detect an electric field of a charge q, we can introduce a test charge q0 and measure the force acting on it.

$\overrightarrow {F}$ = $\frac{1}{4~\pi~ε_0} \frac {q q_0}{r^2} \hat r$

Thus the force exerted per unit charge is:

$\overrightarrow{E}$ = $\frac {\overrightarrow F}{q_0}$ = $\frac {1}{4~\pi~ε_0 } \frac{q}{r^2} \hat r$

Note that the electric field is a vector quantity that is defined at every pint in space, the value of which is dependent only upon the radial distance from q.

The test charge q0 itself has the ability to exert an electric field around it. Hence, to prevent the influence of the test charge, we must ideally make it as small as possible.

Thus,

$\overrightarrow{E}$
=$\lim_{{q_0}\to 0} \frac {\overrightarrow{F}}{q_0}$ = $\lim_{{q_0}\to 0} \frac {1}{4~\pi~ε_0 } \frac{qq_0}{r^2} \hat r × \frac {1}{q_0}$ = $\frac {1}{4~\pi~ε_0 } \frac{q}{r^2} \hat r$

This is the electric field of a point charge. Also, observe that it exhibits spherical symmetry since the electric field has the same magnitude on every point of an imaginary sphere centered around the charge q.

Electric field due to positive and negative charges

Read more on physics calculators and physics formulas at BYJU’S

#### Practise This Question

The electric field in a region is given by
E=(2^i+3^j)×103NC1
The electric flux through a rectangular surface 10cm×20cm held parallel to the y-z plance is