A uniformly charged sphere has a potential on its surface of 410 V. At a radial distance of 20 cm from this surface, the potential is 150 V. The Coulomb constant is 8.99 × 109 N · m2/C2.What is the radius of the sphere? Answer in units of m.
The potential is proportional to 1/R, where R is the distance from the center. You don't need Coulomb's constant for this one. Let R1 be the surface and R2 = 20 cm.
V2/V1 = R1/R2
150/410 = R1/20
R1 = 7.37 cm
To find the radius of the sphere, we can use the equation for electric potential:
V = k * (Q / r)
Where:
V is the electric potential
k is the Coulomb constant (8.99 × 10^9 N · m^2/C^2)
Q is the total charge of the sphere
r is the radial distance from the center of the sphere
We are given the potential at two points: on the surface of the sphere (V = 410 V) and at a radial distance of 20 cm (0.2 m) from the surface (V = 150 V).
Let's start by finding the potential at the surface of the sphere (V = 410 V):
410 V = k * (Q / R)
where R is the radius of the sphere.
Next, let's find the potential at a radial distance of 20 cm (V = 150 V):
150 V = k * (Q / 0.2 m)
Now, we can set up a ratio between the potentials:
(V at the surface) / (V at 20 cm) = (Q / R) / (Q / 0.2 m)
Simplifying the equation:
410 V / 150 V = R / 0.2 m
Now, solve for R:
R = (410 V / 150 V) * 0.2 m
Calculating the value:
R = 0.5466666667 m
Thus, the radius of the uniformly charged sphere is approximately 0.55 m.