A uniformly charged insulating rod of length 19.0 cm is bent into the shape of a semicircle. The rod has a total charge of −8.50 ìC.
Find the electric potential at O, the centre of the semicircle.
To find the electric potential at point O, the center of the semicircle, we can use the principle of superposition. We can consider the uniformly charged insulating rod to be composed of small charged elements, each having charge dq.
First, let's find an expression for the electric potential due to a small charged element dq at point O. We can use the formula for electric potential due to a point charge:
dV = k * dq / r
where dV is the electric potential due to the point charge, k is the electrostatic constant (8.99 × 10^9 N m^2/C^2), dq is the charge of the small element, and r is the distance from the element to point O.
In this case, we need to integrate this expression over the entire length of the rod. Since the rod is bent into the shape of a semicircle, we can express the distance r in terms of the angle θ measured from the positive x-axis:
r = R * sin(θ/2)
where R is the radius of the semicircle and θ ranges from 0 to π.
Now, we need to express dq in terms of the length element dl. Since the rod is uniformly charged, we can write:
dq = λ * dl
where λ is the linear charge density and dl is an infinitesimal length element.
The linear charge density is given by:
λ = Q / L
where Q is the total charge of the rod and L is the length of the rod.
Now, we can substitute these expressions back into the integral:
V = ∫ dV = ∫ k * dq / r = k * ∫ λ * dl / (R * sin(θ/2))
Since the rod is bent into a semicircle, the length element dl can be expressed in terms of the radius R and angle θ as:
dl = R * dθ
Substituting this back into the integral:
V = k * ∫ λ * (R * dθ) / (R * sin(θ/2)) = k * λ * R * ∫ dθ / sin(θ/2)
Now we can calculate the potential at point O by evaluating this integral:
V = k * λ * R * ∫ dθ / sin(θ/2) from 0 to π
Substituting the given values:
V = (8.99 × 10^9 N m^2/C^2) * (8.50 × 10^(-6) C) * (0.190 m) * ∫ dθ / sin(θ/2) from 0 to π
Evaluating the integral:
V = (8.99 × 10^9 N m^2/C^2) * (8.50 × 10^(-6) C) * (0.190 m) * [2 * ln(cosec(θ/2) + cot(θ/2))] from 0 to π
Evaluating this expression will give us the electric potential at point O.