Two point dipoles phat-k- and dfrac-p-2-hat-k- are located -0-0-0- and -1m-0-2m- Find the field the point -1-0-0-displaystylefrac-9p-32pi varepsilon-0-hat-k-displaystylefrac-7p-32pi varepsilon-0-hat-k-displaystylefrac-7p-32pi varepsilon-0-hat-k-none of these

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Toppr Admin · Accepted Answer

The given point is on axis of-xA0-xAF-p2 dipole and at equatorial line of -xAF-p dipole so that field at given point is -xAF-E1-xAF-E2-xAF-E1-2K-p-2-23-Kp8-k-xAF-E2-Kp1-x2212-k-xAF-E1-xAF-E2-78Kp-x2212-k-x2212-7p32-x3C0-x3B5-0-k

Two point dipoles $p^k$ and $\frac{p}{2}^k$ are located at (0,0,0) and (1m,0,2m). Find the field at the point (1,0,0).
$\frac{9 p}{32 π ε_{0}}^k$
$\frac{- 7 p}{32 π ε_{0}}^k$
$\frac{7 p}{32 π ε_{0}}^k$
none of these

The given point is on axis of $\frac{¯ p}{2}$ dipole and at equatorial line of $¯ p$ dipole so that field at given point is $(_{1} +_{2})$
$_{1} = \frac{2 K (p / 2)}{2^{3}} = \frac{K p}{8} (+^k)$
$_{2} = \frac{K p}{1} (-^k)$
$_{1} +_{2} = \frac{7}{8} K p (-^k) = - \frac{7 p}{32 π ε_{0}}^k$

Two point dipoles p^k and p2^k are located at (0,0,0) and (1m,0,2m). Find the field at the point (1,0,0).9p32πε0^k−7p32πε0^k7p32πε0^knone of these

The given point is on axis of ¯p2 dipole and at equatorial line of ¯p dipole so that field at given point is (¯E1+¯E2)¯E1=2K(p/2)23=Kp8(+^k)¯E2=Kp1(−^k)¯E1+¯E2=78Kp(−^k)=−7p32πε0^k

Two point dipoles $p^k$ and $\frac{p}{2}^k$ are located at (0,0,0) and (1m,0,2m). Find the field at the point (1,0,0).
$\frac{9 p}{32 π ε_{0}}^k$
$\frac{- 7 p}{32 π ε_{0}}^k$
$\frac{7 p}{32 π ε_{0}}^k$
none of these

The given point is on axis of $\frac{¯ p}{2}$ dipole and at equatorial line of $¯ p$ dipole so that field at given point is $(_{1} +_{2})$
$_{1} = \frac{2 K (p / 2)}{2^{3}} = \frac{K p}{8} (+^k)$
$_{2} = \frac{K p}{1} (-^k)$
$_{1} +_{2} = \frac{7}{8} K p (-^k) = - \frac{7 p}{32 π ε_{0}}^k$