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Electrostatics question

2020 · Q119
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Electrostatics question

2020 · Q119

NEETPhysicsElectrostaticsMCQ+4 / −1
A short electric dipole has a dipole moment of 16 ×\times× 10-9 Cm. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60∘60^\circ60∘ with the dipole axis is :

(14πε0=9×109Nm2/C2)\left( {{1 \over {4\pi {\varepsilon _0}}} = 9 \times {{10}^9}N{m^2}/{C^2}} \right)(4πε0​1​=9×109Nm2/C2)
  1. A
    200 V
  2. B
    400 V
  3. C
    zero
  4. D
    50 V
View written solutionFree

Correct answer: A

Electric potential due to electric dipole (V)

=14πε0p.cos⁡θr2 = {1 \over {4\pi {\varepsilon _0}}}{{p.\cos \theta } \over {{r^2}}}=4πε0​1​r2p.cosθ​

V=9×109×16×109×cos⁡600.36V = {{9 \times {{10}^9} \times 16 \times {{10}^9} \times \cos 60} \over {0.36}}V=0.369×109×16×109×cos60​

V = 200 V

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