Find the potential at the center of curvature of the (thin) wire shown in the figure. It has a (uniformly distributed) charge per unit length of λ = 1.99 ⋅ 10–8 C/m𝜆 = 1.99 · 10–8 C/m and a radius of curvature of R = 5.71 cm. V
Question
Find the potential at the center of curvature of the (thin) wire shown in the figure. It has a (uniformly distributed) charge per unit length of λ = 1.99 ⋅ 10–8 C/m𝜆 = 1.99 · 10–8 C/m and a radius of curvature of R = 5.71 cm. V
Solution
To find the potential at the center of curvature of the wire, we can use the formula for the electric potential due to a line of charge, which is given by:
V = kλ ∫dl/r
where:
- V is the electric potential,
- k is Coulomb's constant (8.99 x 10^9 N m^2/C^2),
- λ is the charge per unit length,
- dl is an infinitesimal element of the line of charge,
- r is the distance from the charge element to the point where we want to find the potential.
In this case, the wire is a circle with radius R, and we want to find the potential at the center of the circle. Therefore, r = R for all charge elements, and the integral simplifies to:
V = kλ ∫dl/R = kλ (2πR)/R = 2πkλ
Substituting the given values:
V = 2π * (8.99 x 10^9 N m^2/C^2) * (1.99 x 10^-8 C/m) = 355.6 V
So, the potential at the center of curvature of the wire is approximately 355.6 volts.
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