Question

An electric dipole is placed making at an angle $60deg$ with an electric field of strength $1\times {10}^{5}N/C$. It experiences a torque equal to $8\sqrt{3}N-m$. Calculate the charge on the dipole, if it is of length $2cm$.

Answer 1

Torque on dipole, T = $P\times E\times$Sin( ¥ )

Where

P = Dipole Moment = 2qr = $q\times$ ( 2 r )

Here q = Charge on dipole ,

2r = length of dipole = 2cm = 0.02m

E = Electric Field intensity = 1000000 N/C

¥ = angle between P and E = ${60}^0$

Now

T = $P\times E\times Sin$ ( ¥ )

= $q\times \left(2r\right)\times E\times$ Sin ( ¥ )

= $q\times 2r\times E\times$ Sin ( ¥ )

q = $\frac{T}{2r\times E\times Sin\left(\yen \right)}$

= $\frac{8\sqrt{3}}{0.02\times 1000000}\times \left(\frac{\sqrt{3}}{2}\right)$

= $\frac{\left[8\sqrt{3}\right)\times 100\times 2]}{2\times 1000000\times \left(\sqrt{3}\right)}$

= $\frac{8}{10000}$

= $8\times {10}^{-4}C$

= $8\times {10}^{-4}C\times \frac{100}{100}$

= $8\times 100\times \left[\frac{{10}^{-4}}{100}\right]$C

= $800\times {10}^{-6}$ C

= 800 microCoulomb