Question

Two point-charges of $+1.0\times {10}^{-6}C$ and $-1.0\times {10}^{-6}C$ are at a distance $2.0cm$ apart from each other. This electric dipole is situated in a uniform electric field of $1.0\times {10}^{5}V-m^{-1}$. What will be the maximum torque acting on it due to the field? What will be the work done for rotating it through 180 from the equilibrium position?

Answer 1

ATQ, $Q=1\times {10}^{-6}C$

$d=2cm=2\times {10}^{-2}m$

$\overrightarrow{E}={10}^{5}U/m$

$\therefore$ dipole moment, $P=Qd$

$={10}^{-6}\times 2\times {10}^{-2}$

$=2\times {10}^{-8}Cm$

Torque acting on a dipole is given by, $Z=P\times E$

$=PE\sin \theta$

Now, Torque, $Z$ is maximum when $sin\theta =1$

$\Rightarrow \theta ={90}^0$

$\therefore Z_{max}=P\times E\times sin\theta$

$=2\times {10}^{-8}\times {10}^5\times 1$

$=2\times {10}^{-3}Nm$

Work done is given by, $W=PE[{cos\theta }_i-{cos\theta }_F]$

At equilibrium $={\theta }_i={180}^0$

after rotating, ${\theta }_F=0^0$

$\therefore W=-2PE=-4\times {10}^{-3}J$