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The charge per unit length of the four quadrant of the ring is 2l, _2l, l and _l respectively. The electric field at the centre is -
- a)–
- b)
- c)
- d)None
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The charge per unit length of the four quadrant of the ring is 2l, _2l...
Divide in 4 1/4 rings and use result 2Klemda/r in both x and y axis and add by vector rule hope you do calculation own ur own
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The charge per unit length of the four quadrant of the ring is 2l, _2l...
To understand why the electric field intensity changes when the distance is reduced, we need to recall the formula for the electric field intensity (E) due to a point charge (q) at a distance (r) from it. The formula is given by Coulomb's law:
E = k * (q / r^2)
where k is the electrostatic constant.
1. Given Information:
- Initial distance (r1) = 4 meters
- Electric field intensity at this distance (E1) = 200 N/C
2. Initial Electric Field Intensity:
Using the formula, we can substitute the given values and solve for the initial charge (q1):
200 N/C = k * (q1 / (4 meters)^2)
3. Calculating q1:
To find q1, we need to know the value of the electrostatic constant (k). The value of k is approximately 9 x 10^9 N m^2/C^2. Substituting this value into the equation:
200 N/C = (9 x 10^9 N m^2/C^2) * (q1 / 16 m^2)
Simplifying the equation gives:
q1 = (200 N/C) * (16 m^2) / (9 x 10^9 N m^2/C^2)
q1 = 3.56 x 10^-7 C
4. Changing the Distance:
Now, let's consider what happens when the distance is reduced to 2 meters (r2). We want to find the new electric field intensity (E2) at this distance.
Using the same formula and substituting the new values, we have:
E2 = k * (q1 / (2 meters)^2)
E2 = (9 x 10^9 N m^2/C^2) * (3.56 x 10^-7 C / 4 m^2)
E2 = 2.52 x 10^3 N/C
5. Final Answer:
So, when the distance is reduced from 4 meters to 2 meters, the electric field intensity increases to approximately 2520 N/C (or 2.52 x 10^3 N/C).
Therefore, the correct answer is option 'C' (800 N/C).
E = k * (q / r^2)
where k is the electrostatic constant.
1. Given Information:
- Initial distance (r1) = 4 meters
- Electric field intensity at this distance (E1) = 200 N/C
2. Initial Electric Field Intensity:
Using the formula, we can substitute the given values and solve for the initial charge (q1):
200 N/C = k * (q1 / (4 meters)^2)
3. Calculating q1:
To find q1, we need to know the value of the electrostatic constant (k). The value of k is approximately 9 x 10^9 N m^2/C^2. Substituting this value into the equation:
200 N/C = (9 x 10^9 N m^2/C^2) * (q1 / 16 m^2)
Simplifying the equation gives:
q1 = (200 N/C) * (16 m^2) / (9 x 10^9 N m^2/C^2)
q1 = 3.56 x 10^-7 C
4. Changing the Distance:
Now, let's consider what happens when the distance is reduced to 2 meters (r2). We want to find the new electric field intensity (E2) at this distance.
Using the same formula and substituting the new values, we have:
E2 = k * (q1 / (2 meters)^2)
E2 = (9 x 10^9 N m^2/C^2) * (3.56 x 10^-7 C / 4 m^2)
E2 = 2.52 x 10^3 N/C
5. Final Answer:
So, when the distance is reduced from 4 meters to 2 meters, the electric field intensity increases to approximately 2520 N/C (or 2.52 x 10^3 N/C).
Therefore, the correct answer is option 'C' (800 N/C).
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The charge per unit length of the four quadrant of the ring is 2l, _2l, l and _l respectively. The electric field at the centre is - a)–b)c)d)NoneCorrect answer is option 'A'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The charge per unit length of the four quadrant of the ring is 2l, _2l, l and _l respectively. The electric field at the centre is - a)–b)c)d)NoneCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The charge per unit length of the four quadrant of the ring is 2l, _2l, l and _l respectively. The electric field at the centre is - a)–b)c)d)NoneCorrect answer is option 'A'. Can you explain this answer?.
The charge per unit length of the four quadrant of the ring is 2l, _2l, l and _l respectively. The electric field at the centre is - a)–b)c)d)NoneCorrect answer is option 'A'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The charge per unit length of the four quadrant of the ring is 2l, _2l, l and _l respectively. The electric field at the centre is - a)–b)c)d)NoneCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The charge per unit length of the four quadrant of the ring is 2l, _2l, l and _l respectively. The electric field at the centre is - a)–b)c)d)NoneCorrect answer is option 'A'. Can you explain this answer?.
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