Name: Introduction to Electric Charge
Uploaded: 2019-07-19
Duration: 9 min 41 s
Description: In this video, we will study electric charge and its properties and we will also learn about different methods of charging.

Question

Derive an expression for the electric field at any point on the equatorial line of an electric dipole.

Accepted Answer

Consider an electric dipole consisting of charges $- q$ and $+ q$ separated by a distance $2 a$ .

Let $E_{A}$ be the electric field at point $Q$ due to charge $- q$ and $E_{B}$ be the electric field at point $Q$ due to charge $+ q$ where $Q$ is a point on the equatorial line of the dipole whose electric field is to be measured. Let $x$ be the distance between the centre of the dipole and point $Q$ .

Now, resultant electric field at point $Q$ would be,

$E_{Q} = E_{A} + E_{B}$

$E_{A} = \frac{1}{4 π ε_{0}} (\frac{q}{{(A Q)}^{2}})$

$\Rightarrow \frac{1}{4 π ε_{0}} (\frac{q}{x^{2} + a^{2}})$ ( $∵$ According to Pythagoras, $A Q = {(x^{2} + a^{2})}^{2}$ )

$E_{B} = \frac{1}{4 π ε_{0}} (\frac{q}{{(B Q)}^{2}})$

$\Rightarrow \frac{1}{4 π ε_{0}} (\frac{q}{x^{2} + a^{2}})$ ( $∵$ According to Pythagoras, $B P = {(x^{2} + a^{2})}^{2}$ )

Now, according to parallelogram law of vector addition, the resultant electric field would become,

$E_{Q} = \sqrt{{E_{A}}^{2} + {E_{B}}^{2} + 2 E_{A} E_{B} \cos 2 θ}$

Now, since $E_{A} = E_{B}$ we can say, $E_{A} = E_{B} = E'$

$E_{Q} = \sqrt{{E'}^{2} + {E'}^{2} + 2 E' E' \cos 2 θ} \Rightarrow \sqrt{2 {E'}^{2} (1 + \cos 2 θ)} \Rightarrow E' \sqrt{2 (1 + \cos 2 θ)} \Rightarrow E' \sqrt{2 (2 \cos^{2} θ)} E_{Q} = 2 E' \cos θ - - - (i)$

Now, from $△ A O Q$ , $\cos θ = \frac{a}{\sqrt{x^{2} + a^{2}}}$

Also, we know that, $E' = \frac{1}{4 π ε_{0}} (\frac{q}{x^{2} + a^{2}})$

Putting both these values in equation $(i)$ , we get,

$E_{Q} = \frac{2}{4 π ε_{0}} (\frac{q}{x^{2} + a^{2}}) (\frac{a}{\sqrt{x^{2} + a^{2}}}) E_{Q} = \frac{p}{4 π ε_{0} {(x^{2} + a^{2})}^{\frac{3}{2}}}$

Since we know that in dipoles, $x > > > 2 a$ , then $a^{2}$ can be neglected.

$∴ E = \frac{p}{4 π ε_{0} x^{\frac{3}{2}}}$

Embibe Experts Solutions for Chapter: Electric Charges and Fields, Exercise 1: Exercise

Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Electric Charges and Fields, Exercise 1: Exercise

Questions from Embibe Experts Solutions for Chapter: Electric Charges and Fields, Exercise 1: Exercise with Hints & Solutions

Learn and Prepare for any exam you want !