A short electric dipole has a dipole moment of 16 × 10-9 C m. 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 60Â° with the dipole axis is: ( (1/4π ϵo)= 9 × 109 N m2/C2)

Question

A short electric dipole has a dipole moment of 16 × 10-9 C m. 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 60Â° with the dipole axis is: ( (1/4π ϵo)= 9 × 109 N m2/C2) (A) 50 V (B) 200 V (C) 400 V (D) zero

Tardigrade Admin · Accepted Answer

The electric potential due to the dipole, V=(kp cos theta/r2) =(9× 109× 16× 10-9/(0.6)2)× (1/2) V=200 V

$50 V$

$200 V$

$400 V$

zero

The electric potential due to the dipole,
$V = \frac{k p cos t h e t a}{r ^{2}}$
$= \frac{9 \times 1 0 ^{9} \times 16 \times 1 0 ^{- 9}}{( 0.6 ) ^{2}} \times \frac{1}{2}$
$V = 200 V$

Q. A short electric dipole has a dipole moment of 16×10−9Cm. The electric potential due to the dipole at a point at a distance of 0.6m from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis is : (4πϵo​1​=9×109Nm2/C2)

50V

200V

400V

zero

The electric potential due to the dipole, V=r2kpcostheta​ =(0.6)29×109×16×10−9​×21​ V=200V

$50 V$

$200 V$

$400 V$

The electric potential due to the dipole,
$V = \frac{k p cos t h e t a}{r ^{2}}$
$= \frac{9 \times 1 0 ^{9} \times 16 \times 1 0 ^{- 9}}{( 0.6 ) ^{2}} \times \frac{1}{2}$
$V = 200 V$