Question

For a short dipole placed at origin O, the dipole moment P is along the X-axis, as shown in the figure. If the electric potential and electric field at A are V and E respectively, then the correct combination of the electric potential and electric field, respectively, at point B on the Y-axis is given by:

• A. \( \frac{V_0}{4}, \frac{E_0}{4} \)
• B. \( 0, \frac{E_0}{16} \)
• C. \( \frac{V_0}{2}, \frac{E_0}{16} \)
• D. \( \frac{E_0}{8} \)

Hint:

For a short dipole: - The potential along the perpendicular bisector is always zero. - The electric field along the perpendicular bisector follows \( E \propto \frac{1}{r^3} \).

Answer 1

The Correct Option is B

Step 1: Compute the electric potential at point \( B \). The electric potential due to a dipole at any point is given by: \[ V = \frac{1}{4\pi\epsilon_0} \frac{\mathbf{p} \cdot \hat{r}}{r^2}. \] Since point \( B \) is on the perpendicular bisector of the dipole, \( \mathbf{p} \cdot \hat{r} = 0 \), implying: \[ V_B = 0. \] Step 2: Compute the electric field at point \( B \). The magnitude of the electric field along the perpendicular bisector of a dipole is: \[ E = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + d^2)^{3/2}}. \] For small dipole approximation \( d \ll r \), we use: \[ E_B = \frac{E_0}{16}. \] Thus, the answer is \( \boxed{0, \frac{E_0}{16}} \).

Answer 2

Step 1: Recall formulas for electric potential and field due to a short dipole. 

For a short dipole of dipole moment \( p \) at a point with spherical coordinates \( (r, \theta) \): \[ V = \frac{1}{4\pi\varepsilon_0} \frac{p \cos\theta}{r^2} \] and \[ E_r = \frac{1}{4\pi\varepsilon_0} \frac{2p \cos\theta}{r^3}, \quad E_\theta = \frac{1}{4\pi\varepsilon_0} \frac{p \sin\theta}{r^3}. \] The resultant field magnitude is: \[ E = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}\sqrt{1 + 3\cos^2\theta}. \] ---

Step 2: For point \( A \) on the X-axis.

Here, \( \theta = 0^\circ \Rightarrow \cos\theta = 1, \sin\theta = 0. \) Hence: \[ V_A = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^2} = V_0, \] \[ E_A = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3} = E_0. \] ---

Step 3: For point \( B \) on the Y-axis.

Here, \( \theta = 90^\circ \Rightarrow \cos\theta = 0, \sin\theta = 1. \) So the potential: \[ V_B = \frac{1}{4\pi\varepsilon_0}\frac{p\cos 90^\circ}{r^2} = 0. \] and electric field magnitude: \[ E_B = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3} = \frac{E_0}{2}. \] However, since the problem states that the point \( B \) is twice as far from the dipole as \( A \) (from the given figure), the field decreases with the cube of distance: \[ E_B = \frac{E_0}{2^3} = \frac{E_0}{8}. \] But along the equatorial line (Y-axis), the field is half of that compared to the axial value at the same distance, hence: \[ E_B = \frac{E_0}{16}. \] ---

Step 4: Final values at point \( B \):

\[ V_B = 0, \quad E_B = \frac{E_0}{16}. \] 

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Final Answer:

\[ \boxed{V_B = 0,\quad E_B = \frac{E_0}{16}} \]