How do electric field lines depend on the area or the solid angle subtended by the area?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The figure shows a set of electric field lines originating from a point charge $q$.
Consider two small area elements placed at points $R$ and $S$,oriented normal to the field lines.
The number of field lines passing through a given area is proportional to the magnitude of the electric field at that location. The diagram illustrates that the field at $R$ is stronger than at $S$ because the field lines are more densely packed at $R$.
In three dimensions,the solid angle $\Delta \Omega$ subtended by an area element $\Delta S$ at a distance $r$ from the charge is given by $\Delta \Omega = \frac{\Delta S}{r^2}$,which implies $\Delta S = r^2 \Delta \Omega$.
For a fixed solid angle $\Delta \Omega$,the number of radial field lines $n$ passing through the area element is constant.
At two points $P_1$ and $P_2$ at distances $r_1$ and $r_2$ from the charge,the area elements subtending the same solid angle $\Delta \Omega$ are $A_1 = r_1^2 \Delta \Omega$ and $A_2 = r_2^2 \Delta \Omega$,respectively.
The number of field lines $n$ cutting these area elements is the same. Therefore,the number of field lines per unit area (which represents the field strength $E$) is:
$E_1 = \frac{n}{A_1} = \frac{n}{r_1^2 \Delta \Omega}$
$E_2 = \frac{n}{A_2} = \frac{n}{r_2^2 \Delta \Omega}$
Since $n$ and $\Delta \Omega$ are constant,it follows that the strength of the electric field is inversely proportional to the square of the distance,i.e.,$E \propto \frac{1}{r^2}$.

Explore More

Similar Questions

Three positive charges of equal value $q$ are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in:

If charge $q$ is placed on one of the vertex of a cube, then total electric flux passing through the cube is . . . . . . .

The adjoining diagram shows the electric lines of force emerging from a charged body. If the electric fields at $A$ and $B$ are $E_A$ and $E_B$ respectively and the distance between them is $r$,then

$A$ point charge of $+12 \,\mu C$ is at a distance $6 \,cm$ vertically above the centre of a square of side $12 \,cm$ as shown in the figure. The magnitude of the electric flux through the square will be ....... $\times 10^{3} \,Nm^{2}/C$.

The electric field in a region of space is given by $\vec{E} = (5\hat{i} + 2\hat{j}) \text{ N/C}$. Calculate the electric flux through a surface of area $2 \text{ m}^2$ lying in the $YZ$-plane.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo