A charge of 21 nC is uniformly distributed along a straight rod of length 4.3 m that is bent into a circular arc with a radius of 1.7 m. What is the magnitude of the electric field at the center of curvature of the arc?
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page21
To find the magnitude of the electric field at the center of curvature of the arc, we can use the principle of superposition.
The electric field at a point on the arc due to a small element of charge can be calculated using Coulomb's Law:
dE = (k * dq) / r²
Where:
- dE is the electric field due to the small element of charge
- k is the Coulomb's constant (8.99 * 10^9 Nm²/C²)
- dq is the small element of charge
- r is the distance between the small element of charge and the point where we want to calculate the electric field
In this case, we have a uniformly distributed charge along the rod, so we need to divide the rod into infinitesimal charge elements and integrate to find the total electric field at the center of curvature.
First, let's calculate the linear charge density (λ) of the rod.
λ = Q / L
Where:
- λ is the linear charge density
- Q is the total charge along the rod (21 nC = 21 * 10^-9 C)
- L is the length of the rod (4.3 m)
λ = (21 * 10^-9 C) / (4.3 m)
Now, we can consider a small charge element on the rod of length dl at a distance x from the center of curvature.
dq = λ * dl
We need to express x in terms of the angle θ subtended at the center of curvature. Since the rod is bent into a circular arc with a radius of 1.7 m, we can use the relationship:
x = R * θ
Where:
- x is the distance from the center of curvature to the small charge element
- R is the radius of the circular arc
- θ is the angle subtended at the center of curvature
Now, let's replace dq and x in the expression for the electric field:
dE = (k * dq) / x²
dE = (k * λ * dl) / (R * θ)²
To find the total electric field at the center, we need to integrate over the entire length of the rod.
E = ∫dE = ∫[(k * λ * dl) / (R * θ)²]
The bounds of the integral will be from -π to π since we need to cover the entire circular arc.
E = ∫[(k * λ * dl) / (R * θ)²] from -π to π
Simplifying the expression, we have:
E = (k * λ / R²) * ∫(dl / θ²) from -π to π
Now, we can evaluate the integral:
∫(dl / θ²) from -π to π = ∫(R * dθ / (R * θ)²) from -π to π
= R * ∫(dθ / θ²) from -π to π
= R * [(-1/θ)] from -π to π
= R * [(1/π) - (-1/π)]
= 2R/π
Substituting the integral back into the expression for the electric field:
E = (k * λ / R²) * (2R/π)
Finally, we substitute the values for k, λ, and R to calculate the magnitude of the electric field at the center of curvature.
E = (8.99 * 10^9 Nm²/C²) * ((21 * 10^-9 C) / (4.3 m)²) * (2 * 1.7 m / π)