Question

An electric dipole is along a uniform electric field. If it is deflected by ${60^o}$, work done by the agent is $2 \times {10^{( - 19)}}J$. Then the work done by an agent if it is deflected by ${30^o}$ further is :
A. $2.5{\text{ }} \times {10^{( - 19)}}J$
B. $2 \times {10^{( - 19)}}J$
C. $4 \times {10^{( - 19)}}J$
D. $4 \times {10^{( - 16)}}J$

Answer 1

When an electric dipole in an electric field, a torque acts on it. This torque tries to rotate through the angle if dipole is rotated through an angle ${\theta _1}$ to ${\theta _2}$, then work done by external force is given by:
$W = PE(cos\;{\theta _2} - co{s_1})$

Complete step by step answer:
Work done by agent is $2 \times {10^{( - 19)}}J$ when electric dipole deflected by ${60^o}$
${\theta _1} = {60^o}$ and ${\theta _2} = {30^o}$
When an electric dipole in an electric field, a torque acts on it. This torque tries to rotate through the angle if dipole is rotated through an angle ${\theta _1}$ to ${\theta _2}$, then work done by external force is given by:
$W{\text{ }} = - PE{\text{ }}(cos\;{\theta _2} - cos{\theta _1})$
Working on electric dipole when deflected by an angle of ${60^o}$ is given by,
$W_1 = U = - PEcos{60^o} = - 2 \times {10^{( - 19)}}J$.
Now, work done in deflecting he dipole by another ${30^o}$ is given by,
$W_1 = - PEcos{90^o} = 0$
Therefore work done by the agency which deflected dipole by 300 m pore is,
$W = W_2 - W_1 \\\ \implies W = 0 - ( - 2 \times {10^{( - 19)}})W \\\ \implies W = 2 \times {10^{( - 19)}}J \\\$
The work done by an agent if it is deflected by ${30^o}$

So, the correct answer is “Option B”.

Note:
When an electric dipole in an electric field, a torque acts on it.
This torque tries to rotate through the angle if the dipole is rotated through an angle ${\theta _1}$ to ${\theta _2}$.