Question

An electric dipole is along a uniform electric field. If it is deflected by ${{6}}{{{0}}^{{0}}}$, work done by agent is ${{2 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}$. Then the work done by an agent if it is deflected by ${{3}}{{{0}}^{{0}}}$ further is :
A. ${{2}}{{.5 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}$
B. ${{2 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}$
C. ${{4 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}$
D. ${{2 \times 1}}{{{0}}^{{{ - 18}}}}{{J}}$

Answer 1

Hint: 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:
${{W = - PE}}\left( {{{cos}}{{{\theta }}_{{2}}}{{ - cos}}{{{\theta }}_{{1}}}} \right)$

Complete step by step solution: Given:
Work done by agent is ${{2 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}$ when electric dipole deflected by ${{6}}{{{0}}^{{0}}}$ .
 ${{{\theta }}_{{1}}} = {{6}}{{{0}}^{{0}}}$ and ${{{\theta }}_{{2}}} = 3{{{0}}^{{0}}}$
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:
 ${{W = - PE}}\left( {{{cos}}{{{\theta }}_{{2}}}{{ - cos}}{{{\theta }}_{{1}}}} \right)$
Working on electric dipole when deflected by an angle of ${{6}}{{{0}}^{{0}}}$ is given by,
$\
  {{{W}}_{{1}}}{{ = - PEcos6}}{{{0}}^{{0}}} \\
  {{ = - 2 \times 1}}{{{0}}^{{{ - 19}}}}{{J}} \\
$
Now, work done in deflecting the dipole by another ${{3}}{{{0}}^{{0}}}$ is given by,
${{{W}}_{{1}}}{{ = - PEcos9}}{{{0}}^{{0}}}{{ = 0}}$
Therefore work done by the agency which deflected dipole by ${30^ \circ }$ more is,
${{W = }}{{{W}}_{{2}}}{{ - }}{{{W}}_{{1}}}$
 \[ \Rightarrow {{W = 0 - }}\left( {{{ - 2 \times 1}}{{{0}}^{{{ - 19}}}}} \right)\]
$ \Rightarrow {{W = 2 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}$
The work done by an agent if it is deflected by ${{3}}{{{0}}^{{0}}}$ is ${{2 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}$.

Hence, option (B) is the correct answer.

Note: Here we have to see which angle is ${{6}}{{{0}}^{{0}}}$ and which angle is ${{9}}{{{0}}^{{0}}}$ . Also we have to add ${{3}}{{{0}}^{{0}}}$ more to make the angle ${{9}}{{{0}}^{{0}}}$ . So, we have to observe the question carefully as to what angle is given.