SATHEE: Electric Charges and Fields

Electric Charges and Fields - Result Question 24

24. The electric intensity due to a dipole of length $10 c m$ and having a charge of $500 μ C$ , at a point on the axis at a distance $20 c m$ from one of the charges in air, is

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c3eec34ec6b1fad69db54a20ad4b2dca40c2aa54 (a) $6.25 \times 10^{7} N / C$

(b) $9.28 \times 10^{7} N / C$

(d) $20.5 \times 10^{7} N / C$

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Answer:

Correct Answer: 24. (a)

Solution:

(a) Given : Length of the dipole $(2 l) = 10 c m$ $= 0.1 m$ or $l = 0.05 m$

Charge on the dipole $(q) = 500 μ C = 500 \times$ $10^{- 6} C$ and distance of the point on the axis from the mid-point of the dipole $(r) = 20 + 5 = 25 c m =$ $0.25 m$ . We know that the electric field intensity due to dipole on the given point $(E) =$

$\frac{1}{4 π ε_{0}} \times \frac{2 (q .2 l) r}{(r^{2} - l^{2})^{2}}$

$= 9 \times 10^{9} \times \frac{2 (500 \times 10^{- 6} \times 0.1) \times 0.25}{[(0.25)^{2} - (0.05)^{2}]^{2}}$

$= \frac{225 \times 10^{3}}{3.6 \times 10^{- 3}} = 6.25 \times 10^{7} N / C$

The dipole field $E \propto \frac{1}{r^{3}}$ decreases much rapidly as compared to the field of a point charge $E \propto \frac{1}{r^{2}}$

Electric Charges and Fields - Result Question 24

24. The electric intensity due to a dipole of length $10 c m$ and having a charge of $500 μ C$ , at a point on the axis at a distance $20 c m$ from one of the charges in air, is

Answer:

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Electric Charges and Fields - Result Question 24

24. The electric intensity due to a dipole of length 10cm and having a charge of 500μC, at a point on the axis at a distance 20cm from one of the charges in air, is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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24. The electric intensity due to a dipole of length $10 c m$ and having a charge of $500 μ C$ , at a point on the axis at a distance $20 c m$ from one of the charges in air, is