If the magnitude of intensity of electric field at a distance x on axial line and at a distance y on equatorial line on a given dipole are equal, then x: y is

Question

If the magnitude of intensity of electric field at a distance x on axial line and at a distance y on equatorial line on a given dipole are equal, then x: y is (A) 1: 1 (B) 1: √2 (C) 1: 2 (D) √[3]2: 1

Tardigrade Admin · Accepted Answer

E text axial =E text equatorial ⇒ k ⋅ (2 p/x3)=(k ⋅ p/y3) ⇒ (x/y)=(21 / 3/1)=√[3]2: 1

Q. If the magnitude of intensity of electric field at a distance $x$ on axial line and at a distance $y$ on equatorial line on a given dipole are equal, then $x : y$ is

1 : 1

$1 : 2$

1 : 2

$32 : 1$

$E_{axial} = E_{equatorial}$
$\Rightarrow k \cdot \frac{2 p}{x ^{3}} = \frac{k \cdot p}{y ^{3}}$
$\Rightarrow \frac{x}{y} = \frac{2 ^{1/3}}{1} = 32 : 1$

Q. If the magnitude of intensity of electric field at a distance x on axial line and at a distance y on equatorial line on a given dipole are equal, then x:y is

1 : 1

1:2​

1 : 2

32​:1

Eaxial ​=Eequatorial ​ ⇒k⋅x32p​=y3k⋅p​ ⇒yx​=121/3​=32​:1

Q. If the magnitude of intensity of electric field at a distance $x$ on axial line and at a distance $y$ on equatorial line on a given dipole are equal, then $x : y$ is

$1 : 2$

$32 : 1$

$E_{axial} = E_{equatorial}$
$\Rightarrow k \cdot \frac{2 p}{x ^{3}} = \frac{k \cdot p}{y ^{3}}$
$\Rightarrow \frac{x}{y} = \frac{2 ^{1/3}}{1} = 32 : 1$