A solid conducting sphere of radius R is charged such that the magnitude of the electric field at the surface of the sphere is 4.239 ⋅ 105 V/m4.239 · 105 V/m . The potential 28.37 cm from the surface of the sphere is 6.288E5 V. What is the radius R of the sphere?
Question
A solid conducting sphere of radius R is charged such that the magnitude of the electric field at the surface of the sphere is 4.239 ⋅ 105 V/m4.239 · 105 V/m . The potential 28.37 cm from the surface of the sphere is 6.288E5 V. What is the radius R of the sphere?
Solution
The electric field E at the surface of a charged sphere is given by the equation E = kQ/R^2, where k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), Q is the charge on the sphere, and R is the radius of the sphere.
Given that E = 4.239 x 10^5 V/m, we can rearrange the equation to solve for Q:
Q = E * R^2 / k
The electric potential V at a distance r from the center of the sphere is given by the equation V = kQ/r. Given that V = 6.288 x 10^5 V and r = R + 0.2837 m (the radius of the sphere plus the distance from the surface), we can substitute Q from the previous equation into this one to get:
V = k * (E * R^2 / k) / (R + 0.2837)
Solving this equation for R will give us the radius of the sphere.
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