The electric field due to a short electric dipole at a distance r on the axial line from its mid-point is x times the electric field at a distance 2 r on the equatorial line from the mid-point of dipole. Then, the value of x is

Question

The electric field due to a short electric dipole at a distance r on the axial line from its mid-point is x times the electric field at a distance 2 r on the equatorial line from the mid-point of dipole. Then, the value of x is (A) 16 (B) 9 (C) 25 (D) 36

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d3911fa311a8a8df251da92acacb5a0c-.png" /> On axis, field magnitude, E1=(2 k p/r3) On equatorial axis, field magnitude E2=(k p/(2 r)3)=(k p/8 r3) So, E1=16 E2 ⇒ x=16

Q. The electric field due to a short electric dipole at a distance $r$ on the axial line from its mid-point is $x$ times the electric field at a distance $2 r$ on the equatorial line from the mid-point of dipole. Then, the value of $x$ is

16

9

25

36

On axis, field magnitude, $E_{1} = \frac{2 k p}{r ^{3}}$
On equatorial axis, field magnitude
$E_{2} = \frac{k p}{( 2 r ) ^{3}} = \frac{k p}{8 r ^{3}}$
So, $E_{1} = 16 E_{2}$
$\Rightarrow x = 16$

Q. The electric field due to a short electric dipole at a distance r on the axial line from its mid-point is x times the electric field at a distance 2r on the equatorial line from the mid-point of dipole. Then, the value of x is

16

9

25

36

On axis, field magnitude, E1​=r32kp​ On equatorial axis, field magnitude E2​=(2r)3kp​=8r3kp​ So, E1​=16E2​ ⇒x=16

Q. The electric field due to a short electric dipole at a distance $r$ on the axial line from its mid-point is $x$ times the electric field at a distance $2 r$ on the equatorial line from the mid-point of dipole. Then, the value of $x$ is

On axis, field magnitude, $E_{1} = \frac{2 k p}{r ^{3}}$
On equatorial axis, field magnitude
$E_{2} = \frac{k p}{( 2 r ) ^{3}} = \frac{k p}{8 r ^{3}}$
So, $E_{1} = 16 E_{2}$
$\Rightarrow x = 16$