Question

Find the electric dipole moment due to two charges kept at end points of hypotenuse of a right-angled triangle as shown in figure.

• A. $15\text{C-m}$ along $\text{AB}$
• B. $20\text{C-m}$ along $\text{AC}$
• C. $25\text{C-m}$ along $\text{BC}$
• D. $25\text{C-m}$ along $\text{CB}$

Answer 1

The correct option is D $25\text{C-m}$ along $\text{CB}$

We know that, electric dipole moment $\overrightarrow{p}$ is given as

$\overrightarrow{p}=q\left(2\overrightarrow{l}\right)$

Where,
$q=$ magnitude of charge,
$|2\overrightarrow{l}|=$ distance between the charges

And the direction of the dipole moment will be from negative charge to positive charge.

$|2\overrightarrow{l}|=BC=\sqrt{A{B}^2+A{C}^2}$

$\Rightarrow |2\overrightarrow{l}|=\sqrt{3^2+4^2}=5\text{m}$

So, $\overrightarrow{p}=5\times 5$ along $\text{CB}$

$\Rightarrow \overrightarrow{p}=25\text{C-m}$ along $\text{CB}.$

Hence, option $\left(d\right)$ is correct.