The dipolemoment of the given system is-sqrt-3- ql along perpendicular bisector of q - q lineql sqrt-2- along perpendicular bisector of q - q line2 ql along perpendicular bisector of q - q line0

Question

Toppr Admin · Accepted Answer

See the diagram-Dipole moment of a system of charges is given by-x2192-p-x3A3-qi-x2192-ri-x2192-P-x2212-ql-i-ql-i-x2212-2q-x221A-3-2-xD7-l-j-x2212-x221A-3ql-j

The dipolemoment of the given system is:

$\sqrt{3}$ $q l$ along perpendicular bisector of $q - q$ line
$q l$ $\sqrt{2}$ along perpendicular bisector of $q - q$ line
$2 q l$ along perpendicular bisector of $q - q$ line
$0$

See the diagram.
Dipole moment of a system of charges is given by,
$\to p = Σ q_{i}_{i}$
$\to P = - q l^i + q l^i + (- 2 q) (\sqrt{3} / 2 \times l)^j = - \sqrt{3} q l^j$

The dipolemoment of the given system is:√3 ql along perpendicular bisector of q−q lineql √2 along perpendicular bisector of q−q line2 ql along perpendicular bisector of q−q line0

See the diagram.Dipole moment of a system of charges is given by,→p=Σqi→ri→P=−ql^i+ql^i+(−2q)(√3/2×l)^j=−√3ql^j

The dipolemoment of the given system is:

$\sqrt{3}$ $q l$ along perpendicular bisector of $q - q$ line
$q l$ $\sqrt{2}$ along perpendicular bisector of $q - q$ line
$2 q l$ along perpendicular bisector of $q - q$ line
$0$

See the diagram.
Dipole moment of a system of charges is given by,
$\to p = Σ q_{i}_{i}$
$\to P = - q l^i + q l^i + (- 2 q) (\sqrt{3} / 2 \times l)^j = - \sqrt{3} q l^j$