Q.
An electric dipole is along a uniform electric field: If it is deflected by 60∘, work done by an agent is 2×10−19 Then the work done by an agent if it is deflected by 300 further is
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Electrostatic Potential and Capacitance
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Solution:
Working on electric dipole when deflected by an angle of 60∘ is given by, W1=U=−PEcos60∘=−2×10−19J.
Now, work done in deflecting he dipole by another 30∘ is given by, W1=−PEcos90∘=0
Therefore work done by the agency which deflected dipole by 30∘ mpore is, W=W2−W1=0−(−2×10−19)=2×10−19J.