Two point 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

Question

Two point 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 (A) (9 p/32 π ε0) hatk (B) (-7 p/32 π ε0) hatk (C) (7 p /32 π ε0) hat k  (D) none of these

Tardigrade Admin · Accepted Answer

The given point is on axis of ( hatp/2) dipole and at equatorial line of hatp dipole so that field at given point is ( hatE1+ hatE2) vecE1=(2 K(p 2)/23)=(K p/8)(+ hatk) vecE2=(K p/1)(- hatk) vecE1+ vecE2=-(7/8) K p(- hatk)=-(7 p/32 π ε0) hatk

Q. Two point 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

$\frac{9 p}{32 π ε _{0}} \hat{k}$

$\frac{- 7 p}{32 π ε _{0}} \hat{k}$

$\frac{7 p}{32 π ε _{0}} \hat{k}$

none of these

Q. Two point 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​9p​k^

32πε0​−7p​k^

32πε0​7p​k^

none of these

The given point is on axis of 2p^​​ dipole and at equatorial line of p^​ dipole so that field at given point is (E^1​+E^2​) E1​​=232K(p2)​=8Kp​(+k^) E2​​=1Kp​(−k^) E1​​+E2​​=−87​Kp(−k^)=−32πε0​7p​k^

Q. Two point 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

$\frac{9 p}{32 π ε _{0}} \hat{k}$

$\frac{- 7 p}{32 π ε _{0}} \hat{k}$

$\frac{7 p}{32 π ε _{0}} \hat{k}$