Two short dipoles p hatk and (p/2) hatk are located at (0,0,0) and (1 m , 0,2 m ), respectively. The resultant electric field due to the two dipoles at the point (1 m , 0,0) is (-n p/32 π ε0) hatk. Find n.

Question

Two short dipoles p hatk and (p/2) hatk are located at (0,0,0) and (1 m , 0,2 m ), respectively. The resultant electric field due to the two dipoles at the point (1 m , 0,0) is (-n p/32 π ε0) hatk. Find n. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d58eeb64febcb1d21c415bce49df89ed-.png" /> The given point is at axis of ( vecp/2) dipole and at equatorial line of vecp dipole. So, the total field at the given point is vecE=-(k vecp/(1)3)+(2 k( vecp / 2)/(2)3) =k vecp(-1+(1/8))=(-7 vecp/32 π ε0)

Q. Two short dipoles $p \hat{k}$ and $\frac{p}{2} \hat{k}$ are located at $(0, 0, 0)$ and $(1 m, 0, 2 m)$ , respectively. The resultant electric field due to the two dipoles at the point $(1 m, 0, 0)$ is $\frac{- n p}{32 π ε _{0}} \hat{k}$ . Find $n$ .

The given point is at axis of $\frac{p}{2}$ dipole and at equatorial line of $p$ dipole.
So, the total field at the given point is
$E = - \frac{k p}{( 1 ) ^{3}} + \frac{2 k ( p /2 )}{( 2 ) ^{3}}$
$= k p (- 1 + \frac{1}{8}) = \frac{- 7 p}{32 π ε _{0}}$

Q. Two short dipoles pk^ and 2p​k^ are located at (0,0,0) and (1m,0,2m), respectively. The resultant electric field due to the two dipoles at the point (1m,0,0) is 32πε0​−np​k^. Find n.

The given point is at axis of 2p​​ dipole and at equatorial line of p​ dipole. So, the total field at the given point is E=−(1)3kp​​+(2)32k(p​/2)​ =kp​(−1+81​)=32πε0​−7p​​

Q. Two short dipoles $p \hat{k}$ and $\frac{p}{2} \hat{k}$ are located at $(0, 0, 0)$ and $(1 m, 0, 2 m)$ , respectively. The resultant electric field due to the two dipoles at the point $(1 m, 0, 0)$ is $\frac{- n p}{32 π ε _{0}} \hat{k}$ . Find $n$ .

The given point is at axis of $\frac{p}{2}$ dipole and at equatorial line of $p$ dipole.
So, the total field at the given point is
$E = - \frac{k p}{( 1 ) ^{3}} + \frac{2 k ( p /2 )}{( 2 ) ^{3}}$
$= k p (- 1 + \frac{1}{8}) = \frac{- 7 p}{32 π ε _{0}}$