Question

An electric dipole has dipole moment $4\times {10}^{-11}$ Cm and the two charges have magnitudes $4nC$ each. Calculate the electric field due to the dipole $i)$ at a point on the axis at a distance $0.01m$ from one of the charges and $ii)$ at a point on the perpendicular bisector at a distance $0.02m$ from the mid-point.

Answer 1

Dipole moment $=4\times {10}^{-11}㎝$

$9=4nC$ each

distance between charges in dipole =d

$9d=4\times {10}^{-9}\times d_1=4\times {10}^{-11}d{=10}^{-9}m$

a) At axis

$E_{due}to+9=\frac{K_9}{{\left(0.02\right)}^2}{E}_{due}to-9=\frac{-K_9}{{\left(0.02\right)}^2}{E}_{net}=K_9[\frac{1}{{\left(0.01\right)}^2}-\frac{1}{{\left(0.02\right)}^2}]=27\times {10}^{4}n/C$

At perpendicular 0.02m

$E_{net}=2n$

$\left|E_{+9}\right|=\left|E_{-9}\right|$ at point P

given y=0.02

$E_{netP}=2\left|E\right|\cos \theta =\frac{2㎏}{({\left(\frac{d}{2}\right)}^2+y^2)}\cdot \frac{y}{(y^2+{\left(\frac{d}{2}\right)}^2)}=\frac{2\times 9\times {10}^9\times 4\times {10}^{-9}}{({\left(\frac{0.01}{2}\right)}^2+{\left(0.02\right)}^2)}\cdot \frac{0.02}{\sqrt{({\left(0.02\right)}^2+{\left(\frac{0.01}{2}\right)}^2)}}=\frac{8}{{(4.25\times {10}^{-4})}^{3/2}}=9.13\times {10}^{5}N/C$