The electric potential in volts due to an electric dipole of dipole moment 2 × 10-8 coulomb-metre at a distance of 3 m on a line making an angle of 60° with the axis of the dipole is

Question

The electric potential in volts due to an electric dipole of dipole moment 2 × 10-8 coulomb-metre at a distance of 3 m on a line making an angle of 60° with the axis of the dipole is (A) Zero (B) 10 (C) 20 (D) 40

Tardigrade Admin · Accepted Answer

V=(9 × 109 × 2 × 10-8 × (1/2)/9)=10 V

Q. The electric potential in volts due to an electric dipole of dipole moment $2 \times 1 0^{- 8}$ coulomb-metre at a distance of $3 m$ on a line making an angle of $6 0^{\circ}$ with the axis of the dipole is

Zero

10

20

40

$V = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 8} \times \frac{1}{2}}{9} = 10 V$

Q. The electric potential in volts due to an electric dipole of dipole moment 2×10−8 coulomb-metre at a distance of 3m on a line making an angle of 60∘ with the axis of the dipole is

Zero

10

20

40

V=99×109×2×10−8×21​​=10V

Q. The electric potential in volts due to an electric dipole of dipole moment $2 \times 1 0^{- 8}$ coulomb-metre at a distance of $3 m$ on a line making an angle of $6 0^{\circ}$ with the axis of the dipole is

$V = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 8} \times \frac{1}{2}}{9} = 10 V$