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A uniformly charged ring of radius R has a linear charge density λ. What is the electric field at a point on the axis of the ring at a distance x from the center?
- a)E = (kλx)/(R²)
- b)E = (kλx)/(R³)
- c)E = (kλ)/(2πx)
- d)E = (kλ)/(4πx²)
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A uniformly charged ring of radius R has a linear charge density &lamb...
To find the electric field at a point on the axis of a uniformly charged ring, we can use the principle of superposition. We divide the ring into infinitesimally small charge elements and calculate the electric field contribution from each element.
Let's consider a small charge element on the ring with length dθ, where dθ is the differential angle subtended by the element. The linear charge density λ is given by λ = Q / (2πR), where Q is the total charge on the ring and R is the radius of the ring.
The charge element dq can be expressed as dq = λRdθ, as the charge density λ is linear and the length of the charge element is Rdθ.
The electric field contribution from this charge element at a point P on the axis can be calculated using Coulomb's law:
dE = (k * dq) / r²,
where k is the Coulomb's constant, dq is the charge element, and r is the distance from the charge element to point P.
In this case, the distance r can be expressed as r = √(x² + R²), where x is the distance along the axis.
Substituting the expression for dq and r, we have:
dE = (k * λRdθ) / (x² + R²).
To find the total electric field on the axis, we integrate this expression over the entire ring:
E = ∫ dE = ∫ (k * λRdθ) / (x² + R²).
The integral limits are from 0 to 2π, as we need to integrate over the entire ring.
Evaluating this integral, we have:
E = (k * λR) ∫ dθ / (x² + R²)
= (k * λR) * [θ]₀²π / (x² + R²)
= (k * λR * 2π) / (x² + R²).
Therefore, the electric field at a point on the axis of a uniformly charged ring is given by:
E = (k * λR * 2π) / (x² + R²).
Let's consider a small charge element on the ring with length dθ, where dθ is the differential angle subtended by the element. The linear charge density λ is given by λ = Q / (2πR), where Q is the total charge on the ring and R is the radius of the ring.
The charge element dq can be expressed as dq = λRdθ, as the charge density λ is linear and the length of the charge element is Rdθ.
The electric field contribution from this charge element at a point P on the axis can be calculated using Coulomb's law:
dE = (k * dq) / r²,
where k is the Coulomb's constant, dq is the charge element, and r is the distance from the charge element to point P.
In this case, the distance r can be expressed as r = √(x² + R²), where x is the distance along the axis.
Substituting the expression for dq and r, we have:
dE = (k * λRdθ) / (x² + R²).
To find the total electric field on the axis, we integrate this expression over the entire ring:
E = ∫ dE = ∫ (k * λRdθ) / (x² + R²).
The integral limits are from 0 to 2π, as we need to integrate over the entire ring.
Evaluating this integral, we have:
E = (k * λR) ∫ dθ / (x² + R²)
= (k * λR) * [θ]₀²π / (x² + R²)
= (k * λR * 2π) / (x² + R²).
Therefore, the electric field at a point on the axis of a uniformly charged ring is given by:
E = (k * λR * 2π) / (x² + R²).
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Community Answer
A uniformly charged ring of radius R has a linear charge density &lamb...
The electric field at a point on the axis of a uniformly charged ring is given by E = (kλ)/(4πx²), where k is the electrostatic constant, λ is the linear charge density, and x is the distance from the center of the ring.
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A uniformly charged ring of radius R has a linear charge density λ. What is the electric field at a point on the axis of the ring at a distance x from the center?a)E = (kλx)/(R²)b)E = (kλx)/(R³)c)E = (kλ)/(2πx)d)E = (kλ)/(4πx²)Correct answer is option 'D'. Can you explain this answer?
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