Electric Charges and Fields - Result Question 24
24. The electric intensity due to a dipole of length $10 cm$ and having a charge of $500 \mu C$, at a point on the axis at a distance $20 cm$ from one of the charges in air, is
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c3eec34ec6b1fad69db54a20ad4b2dca40c2aa54 (a) $6.25 \times 10^{7} N / C$
(b) $9.28 \times 10^{7} N / C$
(c) $13.1 \times 10^{11} N / C$
(d) $20.5 \times 10^{7} N / C$
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Answer:
Correct Answer: 24. (a)
Solution:
- (a) Given : Length of the dipole $(2 l)=10 cm$ $=0.1 m$ or $l=0.05 m$
Charge on the dipole $(q)=500 \mu C=500 \times$ $10^{-6} C$ and distance of the point on the axis from the mid-point of the dipole $(r)=20+5=25 cm=$ $0.25 m$. We know that the electric field intensity due to dipole on the given point $(E)=$
$\frac{1}{4 \pi \varepsilon_0} \times \frac{2(q .2 l) r}{(r^{2}-l^{2})^{2}}$
$=9 \times 10^{9} \times \frac{2(500 \times 10^{-6} \times 0.1) \times 0.25}{[(0.25)^{2}-(0.05)^{2}]^{2}}$
$=\frac{225 \times 10^{3}}{3.6 \times 10^{-3}}=6.25 \times 10^{7} N / C$
The dipole field $E \propto \frac{1}{r^{3}}$ decreases much rapidly as compared to the field of a point charge $E \propto \frac{1}{r^{2}}$
