Q.36. Two short electric dipoles A and B having dipole moment and respectively are placed with the

Q: What is the correct answer for this exam Physics Q36?

The correct answer is B. Sol. The electric field due to dipole A (axial point) is $E_A = \frac{2kp_1}{x^3}$ along the positive x-axis. The electric field due to dipole B (equatorial point) is $E_B = \frac{kp_2}{x^3}$ along the negative y-axis. The resultant electric field $E_{net}$ makes an angle $\theta = 60^\circ$ with the x-axis. From the diagram, $\tan \theta = \frac{|E_B|}{|E_A|}$. So, $\tan 60^\circ = \frac{kp_2/x^3

Q: Q.36. Two short electric dipoles A and B having dipole moment and respectively are placed with their axis mutually perpendicular as shown in the figure. The resultant electric field at a point x is…

The correct answer is B. Sol. The electric field due to dipole A (axial point) is $E_A = \frac{2kp_1}{x^3}$ along the positive x-axis. The electric field due to dipole B (equatorial point) is $E_B = \frac{kp_2}{x^3}$ along the negative y-axis. The resultant electric field $E_{net}$ makes an angle $\theta = 60^\circ$ with the x-axis. From the diagram, $\tan \theta = \frac{|E_B|}{|E_A|}$. So, $\tan 60^\circ = \frac{kp_2/x^3

Q: What are the options for this this exam Physics question?

A. $\sqrt{3}/2$ · B. $2\sqrt{3}$ · C. $1/\sqrt{3}$ · D. $\sqrt{3}$

Question

Q.36. Two short electric dipoles A and B having dipole moment  and  respectively are placed with their axis mutually perpendicular as shown in the figure. The resultant electric field at a point x is making an angle of 60° with the line joining points O and x. The ratio of the dipole moments  is [Figure: Two perpendicular dipoles and electric field at point x].

FirstInTest · Accepted Answer

Sol. The electric field due to dipole A (axial point) is $E_A = \frac{2kp_1}{x^3}$ along the positive x-axis. The electric field due to dipole B (equatorial point) is $E_B = \frac{kp_2}{x^3}$ along the negative y-axis. The resultant electric field $E_{net}$ makes an angle $	heta = 60^\circ$ with the x-axis. From the diagram, $	an 	heta = \frac{|E_B|}{|E_A|}$. So, $	an 60^\circ = \frac{kp_2/x^3}{2kp_1/x^3} = \frac{p_2}{2p_1}$. $\sqrt{3} = \frac{p_2}{2p_1} \Rightarrow \frac{p_2}{p_1} = 2\sqrt{3}$.

Answer

$\sqrt{3}/2$

Answer

$2\sqrt{3}$

Answer

$1/\sqrt{3}$

Answer

$\sqrt{3}$

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