Question

Two opposite charges each of magnitude $2\mu C$ are $1cm$ apart. Find a electric field at a distance of $5cm$ from the mid-point on axial line of the dipole. Also find the field on equatorial line at the same distance from mid-point.

Answer 1

$\begin{array}{c}{E}_1=E_2=\frac{kq}{r^2}\\ r^2={\left(5\right)}^2+{\left(\frac{1}{2}\right)}^2\\ =25+\frac{1}{4}\\ =\frac{100+1}{4}=\frac{101}{4}c{m}^2\\ E_1=E_2=\frac{k\times 2\times {10}^{-6}}{\frac{101}{4}\times {10}^{-4}}\\ =\frac{9\times {10}^9\times 2\times 4\times {10}^{-6}}{101\times {10}^{-4}}\\ =\frac{72}{101}\times {10}^{7}v/m\\ \cos \theta =\frac{base}{hypotenuse}\\ =\frac{\frac{1}{2}}{\sqrt{{\left(5\right)}^2+{\left(\frac{1}{2}\right)}^2}}=\frac{\frac{1}{2}}{\sqrt{25+\frac{1}{4}}}=\frac{\frac{1}{2}}{\sqrt{\frac{101}{2}}}=\frac{1}{\sqrt{101}}\\ So\\ \overrightarrow{E}=2E_1\cos \theta \\ =2\times \frac{72}{101}\times {10}^7\times \frac{1}{\sqrt{101}}\\ =\frac{144\times {10}^{-7}}{{\left(101\right)}^{\frac{3}{2}}}\end{array}$