For a short dipole placed at origin O, the dipole moment P is along x-axis, as shown in the figure. If the electric potential and electric field at A are and , respectively, then the correct combination of the electric potential and electric field, respe

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Q.
For a short dipole placed at origin O, the dipole moment P is along x-axis, as shown in the figure. If the electric potential and electric field at A are $V_{0}$ and $E_{0}$ , respectively, then the correct combination of the electric potential and electric field, respectively, at point B on the y-axis is given by [2025]

1 $\frac{V_{0}}{2} and \frac{E_{0}}{16}$

2 zero and $\frac{E_{0}}{8}$

3 zero and $\frac{E_{0}}{16}$

4 $V_{0} and \frac{E_{0}}{4}$

Ans.
(3)

At point A (axial)

$| E_{0} |$ $= \frac{2 k_{p}}{r^{3}}, V_{0} = \frac{k_{p}}{r^{2}}$

At point B (equatorial)

$| E |$ $= \frac{k_{p}}{{(2 r)}^{3}} = \frac{E_{0}}{16}$

V = 0.

Solution

1 $\frac{V_{0}}{2} and \frac{E_{0}}{16}$

2 zero and $\frac{E_{0}}{8}$

3 zero and $\frac{E_{0}}{16}$

4 $V_{0} and \frac{E_{0}}{4}$

Ans.
(3)

At point A (axial)

$| E_{0} |$ $= \frac{2 k_{p}}{r^{3}}, V_{0} = \frac{k_{p}}{r^{2}}$

At point B (equatorial)

$| E |$ $= \frac{k_{p}}{{(2 r)}^{3}} = \frac{E_{0}}{16}$

V = 0.

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Solution

1 V02 and E016

2 zero and E08

3 zero and E016

4 V0 and E04

Ans. (3) At point A (axial) |E0|=2kpr3, V0=kpr2 At point B (equatorial) |E|=kp(2r)3=E016 V = 0.

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1 $\frac{V_{0}}{2} and \frac{E_{0}}{16}$

2 zero and $\frac{E_{0}}{8}$

3 zero and $\frac{E_{0}}{16}$

4 $V_{0} and \frac{E_{0}}{4}$

Ans.
(3)

At point A (axial)

$| E_{0} |$ $= \frac{2 k_{p}}{r^{3}}, V_{0} = \frac{k_{p}}{r^{2}}$

At point B (equatorial)

$| E |$ $= \frac{k_{p}}{{(2 r)}^{3}} = \frac{E_{0}}{16}$

V = 0.