The potential at a point P due to an electric dipole is 1.8 × 105 V. If P is at a distance of 50 cm apart from the centre O of the dipole and if CP makes an angle 60° with the positive side of the axial line of the dipole, what is the moment of the dipole?

Question

The potential at a point P due to an electric dipole is 1.8 × 105 V. If P is at a distance of 50 cm apart from the centre O of the dipole and if CP makes an angle 60° with the positive side of the axial line of the dipole, what is the moment of the dipole? (A) 10 C - m (B) 10-3 C - m (C) 10-4 C - m (D) 10-5 C - m

Tardigrade Admin · Accepted Answer

V = (1/4πε0) ( p cosθ/r2) Here, V = 1.8 ×105 V , θ =60° r =50 ×10-2 = 0.5 m ∴ 1.8 ×105 = 9 ×109 ×(p cos60°/(0.5)2) or p = (1.8 ×105 ×0.25 ×2/9 ×109) = 10-5 C - m

Q. The potential at a point P due to an electric dipole is $1.8 \times 1 0^{5} V$ . If P is at a distance of 50 cm apart from the centre O of the dipole and if CP makes an angle 60° with the positive side of the axial line of the dipole, what is the moment of the dipole?

10 C - m

$1 0^{- 3} C - m$

$1 0^{- 4} C - m$

$1 0^{- 5} C - m$

Q. The potential at a point P due to an electric dipole is 1.8×105V. If P is at a distance of 50 cm apart from the centre O of the dipole and if CP makes an angle 60° with the positive side of the axial line of the dipole, what is the moment of the dipole?

10 C - m

10−3C−m

10−4C−m

10−5C−m

V=4πε0​1​r2pcosθ​ Here, V=1.8×105V,θ=60∘ r=50×10−2=0.5m ∴1.8×105=9×109×(0.5)2pcos60∘​ or p=9×1091.8×105×0.25×2​=10−5C−m

Q. The potential at a point P due to an electric dipole is $1.8 \times 1 0^{5} V$ . If P is at a distance of 50 cm apart from the centre O of the dipole and if CP makes an angle 60° with the positive side of the axial line of the dipole, what is the moment of the dipole?

$1 0^{- 3} C - m$

$1 0^{- 4} C - m$

$1 0^{- 5} C - m$