There is a uniform charge distribution given by λ = 3.163 ⋅ 10-8 C/m𝜆 = 3.163 · 10-8 C/m along a thin wire of length L. The wire is then curved into a semicircle that is centered about the origin, so the radius of the semicircle is R = Lπ.𝑅 = 𝐿𝜋. The magnitude of the electric field at the center of the semicircle is 8.788E3 N/C. What is value of L? m
Question
There is a uniform charge distribution given by λ = 3.163 ⋅ 10-8 C/m𝜆 = 3.163 · 10-8 C/m along a thin wire of length L. The wire is then curved into a semicircle that is centered about the origin, so the radius of the semicircle is R = Lπ.𝑅 = 𝐿𝜋. The magnitude of the electric field at the center of the semicircle is 8.788E3 N/C. What is value of L? m
Solution
The electric field at the center of a uniformly charged semicircular arc is given by the formula:
E = λ / (2πε₀R)
where: E is the electric field, λ is the linear charge density, ε₀ is the permittivity of free space (approximately 8.85 x 10^-12 C^2/N*m^2), and R is the radius of the semicircle.
We are given that E = 8.788 x 10^3 N/C and λ = 3.163 x 10^-8 C/m. We are also given that R = L/π. We can substitute these values into the formula and solve for L:
8.788 x 10^3 N/C = (3.163 x 10^-8 C/m) / (2π * 8.85 x 10^-12 C^2/N*m^2 * L/π)
Solving for L gives:
L = (3.163 x 10^-8 C/m) / (2π * 8.85 x 10^-12 C^2/N*m^2 * 8.788 x 10^3 N/C)
After calculating the above expression, we find that L is approximately 0.255 meters.
Similar Questions
+9.70 nC of charge is uniformly distributed along the top half of a thin rod of total length L = 8.00 cm, while -9.70 nC of charge is uniformly distributed along the bottom half of the rod, as shown in the figure. What is the magnitude of the electric field at the dot, a distance r = 21.8 cm from the centre of the rod?
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
A charge per unit length λ = +5.26 μC/m is uniformly distributed along the positive y-axis from y = 0 to y = +a = +0.531 m. A charge per unit length λ = −5.26 μC/m is uniformly distributed along the negative y-axis from y = 0 to y = −a = −0.531 m. What is the magnitude of the electric field at a point on the x-axis a distance x = 0.291 m from the origin? (You may enter your calculation using scientific notation.) N/C
A thin glass rod is bent into a semicircle of radius R = 12.9 cm. A charge Q = 4.95 nC is uniformly distributed along the upper half, and a charge −Q is uniformly distributed along the lower half as shown in the figure. Find the y-component of the electric field at point P, the center of the semicircle. (You may enter your calculation using scientific notation. Include a minus sign if necessary.) N/C
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?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.