A long thin rod is bent into a perfect semicircle of radius 3.00 m. The linear change density of the rod is λ = 6.90 nC/m. How much charge is on a small piece of the rod that subtends an angle Δθ = 0.170 radians?
Question
A long thin rod is bent into a perfect semicircle of radius 3.00 m. The linear change density of the rod is λ = 6.90 nC/m. How much charge is on a small piece of the rod that subtends an angle Δθ = 0.170 radians?
Solution
To solve this problem, we need to use the formula for the arc length of a circle, which is given by s = rθ, where r is the radius of the circle and θ is the angle subtended by the arc at the center of the circle.
Step 1: Convert the angle from radians to degrees. We know that 1 radian is approximately 57.3 degrees. So, 0.170 radians is approximately 0.170 * 57.3 = 9.74 degrees.
Step 2: Calculate the length of the small piece of the rod. Using the formula for the arc length, we get s = rθ = 3.00 m * 0.170 rad = 0.51 m.
Step 3: Calculate the charge on the small piece of the rod. The linear charge density λ is defined as the charge per unit length. So, the charge q on the small piece of the rod is given by q = λs = 6.90 nC/m * 0.51 m = 3.52 nC.
So, the charge on the small piece of the rod that subtends an angle Δθ = 0.170 radians is approximately 3.52 nC.
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