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A substance of dipole moment $4 \times 10^{−8} C\; m$ is placed in an electric field of intensity E. When the direction of the field is changed by an angle of $30^{\circ}$, the heat released by the substance is $2.2\; J.$ What is the value of the electric field intensity E?

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$(B) 1.88 \times 10^8 Vm^{-1}$
Hence B is the correct answer.
answered Jun 13, 2014 by meena.p
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