Question:

Three point charges +q, -q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are

Updated On: Feb 20, 2023
  • $\sqrt{2} \,qa$ along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)
  • qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)
  • $\sqrt{2}\,qa$ along +ve x direction
  • $\sqrt{2}\,qa$ along +ve y direction
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The Correct Option is A

Solution and Explanation

The given charge assembly can be represented using the three co-ordinate axes $x , y$ and $z$ as shown in figure. The charge $-2 q$ is placed at the origin $O$. One $+ q$ charge is place at $( a , o , o )$ and the other $+ q$ charge is placed at $( o , a , o )$. Thus the system has two dipoles along $x$-axis and $y$-axis respectively. As the electric dipole moment is directed from the negative to the positive charge hence the resultant dipole moment will be along OA where coordinates of point $A$ are $( a , a , o )$. The magnitude of each dipole moment, , $p = qa$ So, the magnitude of resultant dipole moment is $P _{ R }=\sqrt{ p ^{2}+ P ^{2}}=\sqrt{( qa )^{2}+( qa )^{2}}=\sqrt{2} qa$
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Concepts Used:

Electric Dipole

An electric dipole is a pair of equal and opposite point charges -q and q, separated by a distance of 2a. The direction from q to -q is said to be the direction in space.

p=q×2a

where,

p denotes the electric dipole moment, pointing from the negative charge to the positive charge.

Force Applied on Electric Dipole