A and B are two points on the axis and the perpendicular bisector, respectively, of an electric dipole. A and B are far away from the dipole and at equal distances from it. The fields at A and B are vecEA and vecEB. Then

Question

A and B are two points on the axis and the perpendicular bisector, respectively, of an electric dipole. A and B are far away from the dipole and at equal distances from it. The fields at A and B are vecEA and vecEB. Then (A)  vecEA= vecEB (B)  vecEA=2 vecEB (C)  vecEA=-2 vecEB (D) | vecEB|=(1/2)| vecEA|, and vecEA is perpendicular to vecEB

Tardigrade Admin · Accepted Answer

Electric field due to dipole at axial position is

E_{axial} = \frac{1}{4 π ε _{0}} \frac{2 p}{r ^{3}}

(along the dipole moment)
Electric field due to dipole at equatorial position

E_{eq} = \frac{1}{4 π ε _{0}} \frac{p}{r ^{3}}

(opposite to dipole moment)
Hence

E_{A} = - 2 E_{B}

Q. $A$ and $B$ are two points on the axis and the perpendicular bisector, respectively, of an electric dipole. $A$ and $B$ are far away from the dipole and at equal distances from it. The fields at $A$ and $B$ are $E_{A}$ and $E_{B}$ . Then

$E_{A} = E_{B}$

$E_{A} = 2 E_{B}$

$E_{A} = - 2 E_{B}$

$∣ ∣ E_{B} ∣ ∣ = \frac{1}{2} ∣ ∣ E_{A} ∣ ∣$ , and $E_{A}$ is perpendicular to $E_{B}$

Electric field due to dipole at axial position is
$E_{axial} = \frac{1}{4 π ε _{0}} \frac{2 p}{r ^{3}}$ (along the dipole moment)
Electric field due to dipole at equatorial position
$E_{eq} = \frac{1}{4 π ε _{0}} \frac{p}{r ^{3}}$ (opposite to dipole moment)
Hence $E_{A} = - 2 E_{B}$

Q. A and B are two points on the axis and the perpendicular bisector, respectively, of an electric dipole. A and B are far away from the dipole and at equal distances from it. The fields at A and B are EA​ and EB​. Then

EA​=EB​

EA​=2EB​

EA​=−2EB​

∣∣​EB​∣∣​=21​∣∣​EA​∣∣​, and EA​ is perpendicular to EB​