A dipole is present in an electrostatic field of magnitude 106N/C. If the work done in rotating it from its position of stable equilibrium to its position of unstable equilibrium is 2×10~23J, then find the magnitude of the dipole moment of this dipole.

10^{- 9}

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10^{- 19}

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10^{- 29}

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10^{- 20}

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Solution

The correct option is C $10^{- 29}$ Cm
Electric field in density $E = 106 N / C$
Work Done $= 2 \times 10^{- 3} J$
$W = p E (cos 0 ° - cos 180 °) 2 p E . . (1) p = \frac{W}{2 E} = \frac{2 \times 10^{- 23}}{2 \times 10^{6}} = 10^{- 29} C m$

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A dipole is present in an electrostatic field of magnitude 106N/C. If the work done in rotating it from its position of stable equilibrium to its position of unstable equilibrium is 2×10~23J, then find the magnitude of the dipole moment of this dipole.

The correct option is C 10−29 CmElectric field in density E=106N/CWork Done =2×10−3JW=pE(cos0°−cos180°)2pE..(1)p=W2E=2×10−232×106=10−29Cm

The correct option is C $10^{- 29}$ Cm
Electric field in density $E = 106 N / C$
Work Done $= 2 \times 10^{- 3} J$
$W = p E (cos 0 ° - cos 180 °) 2 p E . . (1) p = \frac{W}{2 E} = \frac{2 \times 10^{- 23}}{2 \times 10^{6}} = 10^{- 29} C m$