Name: Electric Flux Through a Closed Surface
Uploaded: 2019-07-19
Duration: 10 min 4 s
Description: In this video, let us understand the electric flux through a closed surface of a triangular box using Gauss's law of electrostatics.

Question

Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.

Accepted Answer

Consider an electric dipole consisting of charges $- q$ and $+ q$ which are separated by a distance of $2 a$ with a point $O$ being the centre. Let a point $P$ be at the axial line of dipole whose electric field is to be calculated. Let $r$ be the distance between $P$ and $O$ .

Let $E_{A}$ be the electric field due to charge $- q$ and $E_{B}$ be the electric field due to charge $+ q$ .

$E_{A} = \frac{1}{4 π ε_{0}} (\frac{q}{{(A P)}^{2}}) \Rightarrow \frac{1}{4 π ε_{0}} (\frac{q}{{(a + r)}^{2}})$

$E_{B} = \frac{1}{4 π ε_{0}} (\frac{q}{{(B P)}^{2}}) \Rightarrow \frac{1}{4 π ε_{0}} (\frac{q}{{(r - a)}^{2}})$

Since $E_{A}$ and $E_{B}$ are opposite to each other and $E_{B} > E_{A}$ , then the resultant intensity at $P$ would be,

$E = E_{B} - E_{A}$

$\Rightarrow \frac{1}{4 π ε_{0}} (\frac{q}{{(r - a)}^{2}}) - \frac{1}{4 π ε_{0}} (\frac{q}{{(r + a)}^{2}}) \Rightarrow \frac{q}{4 π ε_{0}} [\frac{1}{{(r - a)}^{2}} - \frac{1}{{(r + a)}^{2}}] \Rightarrow \frac{q}{4 π ε_{0}} [\frac{{(r + a)}^{2} - {(r - a)}^{2}}{{(r^{2} - a^{2})}^{2}}] \Rightarrow \frac{q}{4 π ε_{0}} (\frac{4 r a}{{(r^{2} - a^{2})}^{2}}) \Rightarrow \frac{q}{4 π ε_{0}} (\frac{2 a (2 r)}{{(r^{2} - a^{2})}^{2}}) \Rightarrow \frac{1}{4 π ε_{0}} (\frac{2 p r}{{(r^{2} - a^{2})}^{2}})$

Also, if $r > > > a$ , then,

$E = \frac{1}{4 π ε_{0}} (\frac{2 p}{r^{3}})$ .

Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.

Important Questions on Electric Charges and Fields

Learn and Prepare for any exam you want !