The electric field at X due to dipole P is perpendicular to P , then the angle θ is equal to (figure correction)

Question

The electric field at X due to dipole P is perpendicular to P , then the angle θ is equal to (figure correction) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-qwj2c4znvdo8avt3.jpg" /> (A) 0°  (B) tan- 12 (C) tan- 1√2 (D) 90°

Tardigrade Admin · Accepted Answer

tanα =tan(θ /2), where α +θ =90°

Q. The electric field at $X$ due to dipole $P$ is perpendicular to $P$ , then the angle $θ$ is equal to (figure correction)

$0^{\circ}$

$t a n^{- 1} 2$

$t a n^{- 1} 2$

$9 0^{\circ}$

$t an α = t an \frac{θ}{2},$ where $α + θ = 9 0^{\circ}$

Q. The electric field at X due to dipole P is perpendicular to P , then the angle θ is equal to (figure correction)

0∘

tan−12

tan−12​

90∘

tanα=tan2θ​, where α+θ=90∘

Q. The electric field at $X$ due to dipole $P$ is perpendicular to $P$ , then the angle $θ$ is equal to (figure correction)

$0^{\circ}$

$t a n^{- 1} 2$

$t a n^{- 1} 2$

$9 0^{\circ}$

$t an α = t an \frac{θ}{2},$ where $α + θ = 9 0^{\circ}$