Question

Three charge +2 UC, +2 UC and -4 UC are placed as shown in the figure. The net dipole moment of the charge system is

Answer 1

Given:

Three charge +2 uC, +2 uC and -4 uC are placed as shown in the figure.

To find:

Net dipole moment of the system.

Calculation:

$-4 \: \mu C$ can be written as a sum of $-2\: \mu C \:and\: -2\: \mu C$.

So, one -2 $\mu C$ will contribute to a dipole moment in the y-axis where as the other -2 $\mu C$ will contribute to the dipole moment in the x axis.

Dipole Moment in Y axis ;

$\sf{ = q \times (intercharge \: distance)}$

$\sf{ = 2 \times {10}^{ - 6} \times (1 \times {10}^{ - 3} )}$

$\sf{ = 2 \times {10}^{ - 9} \:Cm }$

Dipole Moment in X axis ;

$\sf{ = q \times (intercharge \: distance)}$

$\sf{ = 2 \times {10}^{ - 6} \times (1 \times {10}^{ - 3} )}$

$\sf{ = 2 \times {10}^{ - 9} \:Cm }$

Net dipole moment will be vector summation of the dipole moment of X and Y axis ;

$| \vec{D}| = \sqrt{ {(2 \times {10}^{ - 9}) }^{2} + {(2 \times {10}^{ - 9}) }^{2} }$

$= > | \vec{D}| = \sqrt{ {2 \times (2 \times {10}^{ - 9}) }^{2} }$

$= > | \vec{D}| = 2 \sqrt{2} \times {10}^{ - 9} \: Cm$

So, final answer is:

$\boxed{ \sf{ \red{ \large{ | \vec{D}| = 2 \sqrt{2} \times {10}^{ - 9} \: Cm }}}}$