Q137 / 180 · KCET 2017 · Physics
Three point charges of \( +2 q,+2 q \) and \( -4 q \) are placed at the corners \( A, B \) and \( C \) of an
equilateral triangle \( \mathrm{ABC} \) of side ' \( \mathrm{x} \) '. The magnitude of the electric dipole moment of this system
is
equilateral triangle \( \mathrm{ABC} \) of side ' \( \mathrm{x} \) '. The magnitude of the electric dipole moment of this system
is
- A\( 2 \) qx
- Correct answerB\( 2 \sqrt{3} \) qx
- C\( 3 \sqrt{2} q x \)
- D\( 3 \mathrm{qx} \)
Solution
Net dipole moment
\[ \begin{array}{l} =\sqrt{4 p^{2}+4 p^{2}+2\left(4 p^{2}\right) \cos 60^{\circ}} \\ =\sqrt{4 p^{2}+4 p^{2}+8 p^{2} \times \frac{1}{2}} \\ =\sqrt{4 p^{2}+4 p^{2}+4 p^{2}}=\sqrt{12 p^{2}}=2 \sqrt{3} p \end{array} \]
Therefore, net dipole moment \( 2 \sqrt{3} q x \)
\[ \begin{array}{l} =\sqrt{4 p^{2}+4 p^{2}+2\left(4 p^{2}\right) \cos 60^{\circ}} \\ =\sqrt{4 p^{2}+4 p^{2}+8 p^{2} \times \frac{1}{2}} \\ =\sqrt{4 p^{2}+4 p^{2}+4 p^{2}}=\sqrt{12 p^{2}}=2 \sqrt{3} p \end{array} \]
Therefore, net dipole moment \( 2 \sqrt{3} q x \)
Hence, the correct answer is option: 'B'.