Question 150: Electrostatics & Capacitance

Question

Two short electric dipoles
A
and
B
having dipole moment $p_1$ and $p_2$ 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 $60^{\circ}$ with the line joining points $O$ and $x$. The ratio of the dipole moments $p_2/p_1$ is _________.

Accepted Answer

Electrostatic field lines originate from positive charges and terminate on negative charges. Unlike magnetic field lines, they do not form closed loops because the electrostatic force is conservative and a charge cannot return to its starting point along a continuous line of force without a change in potential.
Hence, statement (A) is incorrect.
By convention, electric field lines represent the direction of force on a positive test charge. If a charge is positive $(\mathrm{q}>0)$, the field lines point radially outward from the charge.
Hence, the statement B is correct.
Gauss's Law is a fundamental law of electromagnetism that relates electric flux to enclosed charge. It is mathematically derived based on the inverse-square nature of Coulomb's Law ( $\mathrm{F} \propto \frac{1}{\mathrm{r}^2}$ ). So, Gauss-Law is valid only for inverse square force (like gravitational force too).
Hence, the statement C is correct.
The electrostatic field is a conservative field. In a conservative field the work done in moving a charge between two points is independent of the path taken, and the total work done over any closed loop is exactly zero.
Hence, the statement D is Correct.
Coulomb's force is a central force, i.e., it always acts along the line joining the centres of two charges. The torque on the particle $(\vec{	au}=\overrightarrow{\mathrm{r}} 	imes \overrightarrow{\mathrm{F}})$ is zero because the central force $(\overrightarrow{\mathrm{F}})$ is anti-parallel to the position vector, so to make the angular momentum $\left(\vec{	au}=\frac{d \overrightarrow{\mathrm{~L}}}{d \mathrm{t}}ight)$ will be constant. For the constant angular momentum, the motion is in a plane. Therefore, the motion of a particle under Coulomb's force must take in a plane.
Hence, the statement D is correct.
Therefore, statements B, C, D, and E are correct, while statement A is false.
Hence, the correct option is (C).

Answer

$\dfrac{\sqrt{3}}{2}$

Answer

$2\sqrt{3}$

Answer

$\dfrac{1}{\sqrt{3}}$

Answer

$\sqrt{3}$

Electrostatics & Capacitance

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