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Find the potential due to a charged ring of density 2 units with radius 2m and the point at which potential is measured is at a distance of 1m from the ring.
- a)18π
- b)24π
- c)36π
- d)72π
Correct answer is option 'D'. Can you explain this answer?
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Find the potential due to a charged ring of density 2 units with radiu...
Answer: d
Explanation: The potential due to a charged ring is given by λa/2εr, where a = 2m and r = 1m. We get V = 72π volts.
Explanation: The potential due to a charged ring is given by λa/2εr, where a = 2m and r = 1m. We get V = 72π volts.
Most Upvoted Answer
Find the potential due to a charged ring of density 2 units with radiu...
To find the potential due to a charged ring at a point outside of the ring, we can use the formula:
V = (k * Q) / r
Where:
- V is the potential at the point
- k is the electrostatic constant, approximately equal to 9 x 10^9 N m^2/C^2
- Q is the charge of the ring
- r is the distance between the point and the center of the ring
In this case, the charge density of the ring is given as 2 units, and the radius of the ring is 2m. The charge Q can be calculated as:
Q = charge density * circumference of the ring
The circumference of a circle is given by 2 * pi * radius, so:
Q = 2 * (2 * pi * 2) = 16 pi units
The distance between the point and the center of the ring is given as 1m.
Plugging in these values into the formula, we get:
V = (k * Q) / r
= (9 x 10^9 * (16 pi)) / 1
= 144 pi x 10^9 units
The approximate value of this expression is 452,389,342,045 units. Therefore, the potential due to the charged ring at a distance of 1m from the ring is approximately 452,389,342,045 units.
V = (k * Q) / r
Where:
- V is the potential at the point
- k is the electrostatic constant, approximately equal to 9 x 10^9 N m^2/C^2
- Q is the charge of the ring
- r is the distance between the point and the center of the ring
In this case, the charge density of the ring is given as 2 units, and the radius of the ring is 2m. The charge Q can be calculated as:
Q = charge density * circumference of the ring
The circumference of a circle is given by 2 * pi * radius, so:
Q = 2 * (2 * pi * 2) = 16 pi units
The distance between the point and the center of the ring is given as 1m.
Plugging in these values into the formula, we get:
V = (k * Q) / r
= (9 x 10^9 * (16 pi)) / 1
= 144 pi x 10^9 units
The approximate value of this expression is 452,389,342,045 units. Therefore, the potential due to the charged ring at a distance of 1m from the ring is approximately 452,389,342,045 units.
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Find the potential due to a charged ring of density 2 units with radius 2m and the point at which potential is measured is at a distance of 1m from the ring.a)18πb)24πc)36πd)72πCorrect answer is option 'D'. Can you explain this answer?
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