Question:
The ratio of electric fields on the axis and at equator of an electric dipole will be
The ratio of electric fields on the axis and at equator of an electric dipole will be
Updated On: Jun 7, 2024
- $ 1:1 $
- $ 2:1 $
- $ 4:1 $
- $ 1:4 $
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The Correct Option is B
Solution and Explanation
The electric field on the axis of a dipole of length small compared to point where electric field is to be determined is
$E_{A}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 p}{r^{3}}$ ... (i)
where $p$ is electric dipole moment.
Intensity at a point on the equatorial line of a dipole is
$E_{E}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{p}{r^{3}} N / C$ ...(ii)
From Eqs. (i) and (ii),
we get $\frac{E_{A}}{E_{E}} =\frac{2 p}{r^{3}} \times \frac{r^{3}}{p}=\frac{2}{1}$
$E_{A}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 p}{r^{3}}$ ... (i)
where $p$ is electric dipole moment.
Intensity at a point on the equatorial line of a dipole is
$E_{E}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{p}{r^{3}} N / C$ ...(ii)
From Eqs. (i) and (ii),
we get $\frac{E_{A}}{E_{E}} =\frac{2 p}{r^{3}} \times \frac{r^{3}}{p}=\frac{2}{1}$
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).