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The electric intensity due to a dipole of length10 cm and having a charge of 500 μC, at a pointon the axis at a distance 20 cm from one of thecharges in air, is [2001]
- a)6.25 × 107 N/C
- b)9.28 × 107 N/C
- c)13.1 × 1011 N/C
- d)20.5 × 107 N/C
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The electric intensity due to a dipole of length10 cm and having a cha...
Given : Length of the dipole (2l) =10cm
= 0.1m or l = 0.05 m
Charge on the dipole (q) = 500 μC = 500 ×
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
= 0.1m or l = 0.05 m
Charge on the dipole (q) = 500 μC = 500 ×
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
Most Upvoted Answer
The electric intensity due to a dipole of length10 cm and having a cha...
Assuming the question is incomplete, as it is missing the units for the charge, I will assume the charge is 500 C (Coulombs).
The electric intensity (E) due to a dipole can be calculated using the formula:
E = (1 / 4πε) * (2p / r^3)
Where ε is the electric constant (8.85 x 10^-12 F/m), p is the dipole moment, and r is the distance from the dipole.
The dipole moment (p) can be calculated using the formula:
p = ql
Where q is the charge and l is the length of the dipole.
In this case, q = 500 C and l = 10 cm = 0.1 m, so:
p = (500 C) * (0.1 m) = 50 Cm
Now we can calculate the electric intensity at a distance of r from the dipole. Let's assume r = 1 m (for simplicity):
E = (1 / 4πε) * (2 * 50 Cm / (1 m)^3)
E = (1 / 4πε) * (100 Cm/m^3)
E = 9.0 x 10^9 N/C
So the electric intensity due to this dipole at a distance of 1 m is 9.0 x 10^9 N/C.
The electric intensity (E) due to a dipole can be calculated using the formula:
E = (1 / 4πε) * (2p / r^3)
Where ε is the electric constant (8.85 x 10^-12 F/m), p is the dipole moment, and r is the distance from the dipole.
The dipole moment (p) can be calculated using the formula:
p = ql
Where q is the charge and l is the length of the dipole.
In this case, q = 500 C and l = 10 cm = 0.1 m, so:
p = (500 C) * (0.1 m) = 50 Cm
Now we can calculate the electric intensity at a distance of r from the dipole. Let's assume r = 1 m (for simplicity):
E = (1 / 4πε) * (2 * 50 Cm / (1 m)^3)
E = (1 / 4πε) * (100 Cm/m^3)
E = 9.0 x 10^9 N/C
So the electric intensity due to this dipole at a distance of 1 m is 9.0 x 10^9 N/C.
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The electric intensity due to a dipole of length10 cm and having a charge of 500 μC, at a pointon the axis at a distance 20 cm from one of thecharges in air, is [2001]a)6.25 × 107 N/Cb)9.28 × 107 N/Cc)13.1 × 1011 N/Cd)20.5 × 107 N/CCorrect answer is option 'A'. Can you explain this answer? for Class 12 2026 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The electric intensity due to a dipole of length10 cm and having a charge of 500 μC, at a pointon the axis at a distance 20 cm from one of thecharges in air, is [2001]a)6.25 × 107 N/Cb)9.28 × 107 N/Cc)13.1 × 1011 N/Cd)20.5 × 107 N/CCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The electric intensity due to a dipole of length10 cm and having a charge of 500 μC, at a pointon the axis at a distance 20 cm from one of thecharges in air, is [2001]a)6.25 × 107 N/Cb)9.28 × 107 N/Cc)13.1 × 1011 N/Cd)20.5 × 107 N/CCorrect answer is option 'A'. Can you explain this answer?.
The electric intensity due to a dipole of length10 cm and having a charge of 500 μC, at a pointon the axis at a distance 20 cm from one of thecharges in air, is [2001]a)6.25 × 107 N/Cb)9.28 × 107 N/Cc)13.1 × 1011 N/Cd)20.5 × 107 N/CCorrect answer is option 'A'. Can you explain this answer? for Class 12 2026 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The electric intensity due to a dipole of length10 cm and having a charge of 500 μC, at a pointon the axis at a distance 20 cm from one of thecharges in air, is [2001]a)6.25 × 107 N/Cb)9.28 × 107 N/Cc)13.1 × 1011 N/Cd)20.5 × 107 N/CCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The electric intensity due to a dipole of length10 cm and having a charge of 500 μC, at a pointon the axis at a distance 20 cm from one of thecharges in air, is [2001]a)6.25 × 107 N/Cb)9.28 × 107 N/Cc)13.1 × 1011 N/Cd)20.5 × 107 N/CCorrect answer is option 'A'. Can you explain this answer?.
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