Question

If the dipole of moment $2.57\times {10}^{-17}C.m$ is placed into an electric field of magnitude $3.0\times {10}^{4}N/C$ such that the fields lines are aligned at ${30}^o$ with the line joining the dipole, what torque will act on the dipole?

• A. $7.7\times {10}^{-13}Nm$
• B. $3.855\times {10}^{-13}Nm$
• C. $3.855\times {10}^{-15}Nm$
• D. $7.7\times {10}^{-15}Nm$

Answer 1

The correct option is B $3.855\times {10}^{-13}Nm$

Given, dipole moment, $p=2.57\times {10}^{-17}C.m$ and electric field , $E=3\times {10}^{4}N/C$
The torque , $\overrightarrow{\tau }=\overrightarrow{p}\times \overrightarrow{E}$
or $\tau =pE\sin 30$

$=2.57\times {10}^{-17}\times 3\times {10}^4\times 0.5$

$=3.855\times {10}^{-13}N.m$