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Question: An electric dipole is along a uniform electric field. If it is deflected by \({{6}}{{{0}}^{{0}}}\), ...

An electric dipole is along a uniform electric field. If it is deflected by 600{{6}}{{{0}}^{{0}}}, work done by agent is 2×10−19J{{2 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}. Then the work done by an agent if it is deflected by 300{{3}}{{{0}}^{{0}}} further is :
A. 2.5×10−19J{{2}}{{.5 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}
B. 2×10−19J{{2 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}
C. 4×10−19J{{4 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}
D. 2×10−18J{{2 \times 1}}{{{0}}^{{{ - 18}}}}{{J}}

Explanation

Solution

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 θ1{{{\theta }}_{{1}}} to θ2{{{\theta }}_{{2}}} , then work done by external force is given by:
W=−PE(cosθ2−cosθ1){{W = - PE}}\left( {{{cos}}{{{\theta }}_{{2}}}{{ - cos}}{{{\theta }}_{{1}}}} \right)

Complete step by step solution: Given:
Work done by agent is 2×10−19J{{2 \times 1}}{{{0}}^{{{ - 19}}}}{{J}} when electric dipole deflected by 600{{6}}{{{0}}^{{0}}} .
θ1=600{{{\theta }}_{{1}}} = {{6}}{{{0}}^{{0}}} and θ2=300{{{\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 θ1{{{\theta }}_{{1}}} to θ2{{{\theta }}_{{2}}} , then work done by external force is given by:
W=−PE(cosθ2−cosθ1){{W = - PE}}\left( {{{cos}}{{{\theta }}_{{2}}}{{ - cos}}{{{\theta }}_{{1}}}} \right)
Working on electric dipole when deflected by an angle of 600{{6}}{{{0}}^{{0}}} is given by,
 W1=−PEcos600 =−2×10−19J \ {{{W}}_{{1}}}{{ = - PEcos6}}{{{0}}^{{0}}} \\\ {{ = - 2 \times 1}}{{{0}}^{{{ - 19}}}}{{J}} \\\
Now, work done in deflecting the dipole by another 300{{3}}{{{0}}^{{0}}} is given by,
W1=−PEcos900=0{{{W}}_{{1}}}{{ = - PEcos9}}{{{0}}^{{0}}}{{ = 0}}
Therefore work done by the agency which deflected dipole by 30∘{30^ \circ } more is,
W=W2−W1{{W = }}{{{W}}_{{2}}}{{ - }}{{{W}}_{{1}}}
⇒W=0−(−2×10−19)\Rightarrow {{W = 0 - }}\left( {{{ - 2 \times 1}}{{{0}}^{{{ - 19}}}}} \right)
⇒W=2×10−19J\Rightarrow {{W = 2 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}
The work done by an agent if it is deflected by 300{{3}}{{{0}}^{{0}}} is 2×10−19J{{2 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}.

Hence, option (B) is the correct answer.

Note: Here we have to see which angle is 600{{6}}{{{0}}^{{0}}} and which angle is 900{{9}}{{{0}}^{{0}}} . Also we have to add 300{{3}}{{{0}}^{{0}}} more to make the angle 900{{9}}{{{0}}^{{0}}} . So, we have to observe the question carefully as to what angle is given.