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 centre of the semicircle, we can use the principle of superposition. Since the rod is uniformly charged, we can consider it as a collection of small charged elements, each contributing to the electric potential at the centre.
To do this, we will use the formula for electric potential due to a point charge:
V = k * (q / r)
where V is the electric potential, k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), q is the charge, and r is the distance between the charge and the point where we want to find the electric potential.
We will divide the rod into small elements and calculate the electric potential due to each element. Then we will sum up the contributions from all the elements to get the total electric potential at point O.
First, we need to calculate the charge density of the rod.
Charge density (ρ) = total charge / length of the rod
ρ = (-8.50 μC) / (0.19 m)
Now, to calculate the electric potential at point O, we will integrate the contributions from all the small elements.
V = ∫ [k * (dq / r)]
= ∫ [k * (ρ * dx / r)]
In a semicircle, the distance r from the centre O to any point on the rod is constant and is equal to the radius of the semicircle.
Let's assume the radius of the semicircle is R.
Using this, we can rewrite the equation as:
V = k * ρ / R ∫ dx
Since the rod is bent into the shape of a semicircle, the range of integration for x is from -R to R.
Now, let's integrate:
V = k * ρ / R ∫ dx
= k * ρ / R * ∫ dx
= k * ρ / R * [x] evaluated from -R to R.
Evaluating at the limits:
V = k * ρ / R * (R - (-R))
= 2k * ρ
Now, substitute the values:
V = 2 * (9 x 10^9 Nm^2/C^2) * [(-8.50 x 10^-6 C) / (0.19 m)]
Solving this equation will give you the value of the electric potential at point O.