Three charges q, -2q and q are placed at (x=0 | vedclass

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(A) The system consists of a charge $-2q$ at the origin $(0,0,0)$. This can be treated as two separate dipoles of charge $q$ and $-q$ each.One dipole

Vedclass · Accepted Answer

$\sqrt{2}qa$,directed from $(x=0, y=0, z=0)$ to $(x=a, y=a, z=0)$

Vedclass · Answer

(A) The system consists of a charge $-2q$ at the origin $(0,0,0)$. This can be treated as two separate dipoles of charge $q$ and $-q$ each.One dipole is formed by $-q$ at $(0,0,0)$ and $+q$ at $(a,0,0)$,with dipole moment $\vec{p}_1 = qa \hat{i}$.The second dipole is formed by $-q$ at $(0,0,0)$ and $+q$ at $(0,a,0)$,with dipole moment $\vec{p}_2 = qa \hat{j}$.The resultant dipole moment is $\vec{p}_{net} = \vec{p}_1 + \vec{p}_2 = qa \hat{i} + qa \hat{j}$.The magnitude is $|\vec{p}_{net}| = \sqrt{(qa)^2 + (qa)^2} = \sqrt{2}qa$.The direction is along the vector $\hat{i} + \hat{j}$,which points from the origin $(0,0,0)$ to the point $(a,a,0)$.

Three charges $q, -2q$ and $q$ are placed at $(x=0, y=a, z=0)$,$(x=0, y=0, z=0)$ and $(x=a, y=0, z=0)$ respectively. What is the resultant electric dipole moment?

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