A solid conducting sphere (radius = 5.0 cm) has a charge of 0.25 nC distributed uniformly on its surface.If point A is located at the centre of the sphere and point B is 15 cm from the centre, what is the magnitude of the electric potential difference between these two points?

Question

Quizplus · Accepted Answer

We can start by using the equation for electric potential due to a point charge:
V = kq/r
where V is the electric potential, k is Coulomb's constant (9.0 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge.
Since the sphere has a uniform surface charge density, we can treat it as a point charge at its center, so we can use the same equation to find the electric potential at point A (the center of the sphere):
V(A) = kq/R
where R is the radius of the sphere.
Plugging in the given values, we get:
V(A) = (9.0 x 10^9 Nm^2/C^2)(0.25 x 10^-9 C)/(0.05 m) = 45 V
To find the electric potential at point B, we need to take into account the contribution from both the sphere and the charge that is now closer to point B than to point A. Since the charge distribution on the sphere is uniform, we can treat it as a point charge located at its center, and the potential due to this charge at point B is:
V(sphere) = kq/DB
where DB is the distance from the center of the sphere to point B. Substituting the given values, we get:
V(sphere) = (9.0 x 10^9 Nm^2/C^2)(0.25 x 10^-9 C)/(0.15 m) = 15 V
The potential due to the additional charge is:
V(additional) = kq/DB
where q is the charge and DB is the distance from the charge to point B. Since the charge is positive and point B is outside the charge, the potential is positive:
V(additional) = (9.0 x 10^9 Nm^2/C^2)(q)/(0.15 m)
We don't know the value of q, but we do know that the total charge on the sphere is 0.25 nC. Since the charge is uniformly distributed, we can find the charge per unit area on the sphere:
σ = q/A
where A is the surface area of the sphere. Substituting the given values, we get:
σ = (0.25 x 10^-9 C)/(4π(0.05 m)^2) = 398 nC/m^2
The area of the portion of the sphere that is closer to point B than to point A is a spherical cap with base radius 0.15 m and height (from the center of the sphere) of 0.05 m. The surface area of a spherical cap is:
A(cap) = 2πRh
where R is the radius of the sphere and h is the height of the cap. Substituting the given values, we get:
A(cap) = 2π(0.05 m)(0.05 m) = 0.0157 m^2
The charge on this portion of the sphere is:
q = σA(cap) = (398 nC/m^2)(0.0157 m^2) = 6.25 nC
Plugging in the values for q and DB, we get:
V(additional) = (9.0 x 10^9 Nm^2/C^2)(6.25 x 10^-9 C)/(0.15 m) = 375 V
The total potential at point B is the sum of the potentials due to the sphere and the additional charge:
V(B) = V(sphere) + V(additional) = 15 V + 375 V = 390 V
The potential difference between point A and point B is:
ΔV = V(B) - V(A) = 390 V - 45 V = 345 V
Rounding to the nearest tens place, we get:
ΔV ≈ 350 V
The closest answer choice is B) 30 V. This is due to a calculation error, the correct answer is not given.

A Solid Conducting Sphere (Radius = 5