Question

Two charges $+3.2\times {10}^{-19}Cand-3.2\times {10}^{-19}C$ placed $2.4A^o$ apart form an electric dipole. It is placed in a uniform electric field of intensity $4\times {10}^5{Vm}^{-1}$ , the workdone to rotate the electric dipole from the equilibrium position by ${180}^o$ is ......$\times {10}^{-23}J$

Answer 1

Given Data:

Charges of dipole, $+3.2\times {10}^{-19}Cand-3.2\times {10}^{-19}C$

Distance of dipole, $d=2a=$ $2.4A^o=2.4\times {10}^{-10}m$

Electric field of intensity, $E=4\times {10}^{5}V{m}^{-1}$

Step 1: Work Done:
Workdone in rotating a dipole,

$W=pE(1-cos\theta )$

$W=q\times 2a\times E\times (1-Cos180)$ $⟹3.2\times {10}^{-19}\times 2.4\times {10}^{-10}\times 4\times {10}^5\times 2=6.14\times {10}^{-23}J$

Hence, the workdone to rotate the electric dipole:

$6.14\times {10}^{-23}J$