Question

The electric intensity due to a dipole of length $10cm$ and having a charge of $500\mu C,$ at a point on the axis at a distance $20cm$ from one of the charges in air, is

• A. $6.25\times {10}^{7}N/C$
• B. $9.28\times {10}^{7}N/C$
• C. $13.1\times {11}^{11}N/C$
• D. $20.5\times {10}^{7}N/C$

Answer 1

The correct option is A $6.25\times {10}^{7}N/C$

$\text{Step 1: Dipole moment of the dipole}$

Dipole moment is given as: $p=q\times 2l$

Here $l=5cm=0.05m$

$p=500\times {10}^{-6}C\times 0.1m$

$p=5\times {10}^{-5}Cm$

It's direction is from $-q$ to $q$.

$\text{Step 2: Electric field due to dipole moment at point P}$

The point say $P$ is at a distance $20cm$ from charge $q$ as shown in figure. [Refer image $2$]

Electric field due to dipole on the axis is given as:

$E=\frac{1}{4\pi {ϵ}_0}\frac{2pr}{{(r^2-l^2)}^2}$ $....\left(1\right)$

where $r$ is the distance of point $P$ from the center of dipole, i.e. $0.25cm$

$\text{Step 3: Substitute the values}$

From equation $\left(1\right)$

$E=\frac{9\times {10}^9\times 2\times 5\times {10}^{-5}\times 0.25}{\left(\right(0.25{)}^2-(0.05{)}^2{)}^2}$

$=6.25\times {10}^{7}N/C$

Hence, Option(A) is correct.

[Note- The electric field is in the same direction as dipole moment.]