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Physics Question on electrostatic potential and capacitance

An electric dipole is along a uniform electric field: If it is deflected by 60∘60^\circ, work done by an agent is 2×10−192 \times 10^{-19} Then the work done by an agent if it is deflected by 30030^0 further is

A

2.5×10−192.5 \times 10^{-19} J

B

2×10−192 \times 10^{-19} J

C

4×10−194 \times 10^{-19} J

D

2×10−162\times 10^{-16} J

Answer

2×10−192 \times 10^{-19} J

Explanation

Solution

Working on electric dipole when deflected by an angle of 60∘60^{\circ} is given by, W1=U=−PEcos⁡60∘=−2×10−19J.W _{1}= U =- PE \cos 60^{\circ}=-2 \times 10^{-19} J . Now, work done in deflecting he dipole by another 30∘30^{\circ} is given by, W1=−PEcos⁡90∘=0W _{1}=- PE \cos 90^{\circ}=0 Therefore work done by the agency which deflected dipole by 30∘30^{\circ} mpore is, W=W2−W1=0−(−2×10−19)=2×10−19J.W = W _{2}- W _{1}=0-\left(-2 \times 10^{-19}\right)=2 \times 10^{-19} J .