Question 163: Electrostatics & Capacitance

Question

Two shorts dipoles $(A, B), A$ having charges $\pm 2 \mu \mathrm{C}$ and length 1 cm and $B$ having charges $\pm 4 \mu \mathrm{C}$ and length 1 cm are placed with their centres 80 cm apart as shown in the figure. The electric field at a point $P$, equi-distant from the centres of both dipoles is $\_\_\_\_$ N/C.

Accepted Answer

The dipole moment is given by $\mathrm{p}=\mathrm{q} 	imes 2 \mathrm{l}$.
For Dipole A :
$\mathrm{q}_{\mathrm{A}}=2 	imes 10^{-6} \mathrm{C}, 2 \mathrm{l}=1 \mathrm{~cm}=0.01 \mathrm{~m}$.
$$ \mathrm{p}_{\mathrm{A}}=\left(2 	imes 10^{-6}ight)(0.01)=2 	imes 10^{-8} \mathrm{C} \cdot \mathrm{~m} $$
For Dipole B :
$\mathrm{q}_{\mathrm{B}}=4 	imes 10^{-6} \mathrm{C}, 2 \mathrm{l}=1 \mathrm{~cm}=0.01 \mathrm{~m}$.
$$ p_B=\left(4 	imes 10^{-6}ight)(0.01)=4 	imes 10^{-8} \mathrm{C} \cdot \mathrm{~m} $$
The distance between the centres is 80 cm .
Point P is equi-distant from both, so it is at the midpoint : $\mathrm{r}=40 \mathrm{~cm}=0.4 \mathrm{~m}$.
Point P is on the axial line of dipole A and on the equatorial line of dipole B .
Field due to A (Axial) :
$$ \mathrm{E}_{\mathrm{A}}=\frac{1}{4 \pi \epsilon_0} \frac{2 \mathrm{p}_{\mathrm{A}}}{\mathrm{r}^3}=\frac{\left(9 	imes 10^9ight) 2\left(2 	imes 10^{-8}ight)}{(0.4)^3}=\frac{360}{0.064}=5625 \mathrm{~N} / \mathrm{C} $$
It is along the axis of dipole A (horizontal)
Field due to B (Equatorial) :
$$ \mathrm{E}_{\mathrm{B}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{p}_{\mathrm{B}}}{\mathrm{r}^3}=\frac{\left(9 	imes 10^9ight) 4 	imes 10^{-8}}{(0.4)^3}=\frac{360}{0.064}=5625 \mathrm{~N} / \mathrm{C} $$
It acts opposite to the direction of dipole B (vertical).
Since the vectors are perpendicular and equal in magnitude $\left(\mathrm{E}_{\mathrm{A}}=\mathrm{E}_{\mathrm{B}}=\mathrm{E}ight)$ :
$$ \mathrm{E}_{\mathrm{net}}=\sqrt{\mathrm{E}_{\mathrm{A}}^2+\mathrm{E}_{\mathrm{B}}^2}=\mathrm{E} \sqrt{2} $$
$$ \mathrm{E}_{\mathrm{net}}=5625 \sqrt{2} \frac{\mathrm{~N}}{\mathrm{C}}=0.5625 \sqrt{2} 	imes 10^4=\frac{9}{16} \sqrt{2} 	imes 10^4 \frac{\mathrm{~N}}{\mathrm{C}} $$
So, net electric field at point P is, $\mathrm{E}_{	ext {net }}=\frac{9}{16} \sqrt{2} 	imes 10^4 \mathrm{~N} / \mathrm{C}$.
Hence, the correct option is (B).

Answer

$\frac{9}{16} \sqrt{2} 	imes 10^5$

Answer

$\frac{9}{16} \sqrt{2} 	imes 10^4$

Answer

$9 \sqrt{2} 	imes 10^4$

Answer

$4.5 \sqrt{2} 	imes 10^4$

Electrostatics & Capacitance

Two shorts dipoles (A, B), A having charges \pm 2 \mu \mathrm{C} and length 1 cm and B having charges \pm 4 \mu \mathrm{C} and length 1 cm are placed with their centres 80 cm apart as shown in the figure. The electric field at a point P, equi-distant from the centres of both dipoles is \_\_\_\_ N/C.

Choose the correct answer: