Question

A classroom Van de Graaff generator accumulates a charge of \(1.00 \cdot 10^{-6} \mathrm{C}\) on its spherical conductor, which has a radius of \(10.0 \mathrm{~cm}\) and stands on an insulating column. Neglecting the effects of the generator base or any other objects or fields, find the potential at the surface of the sphere. Assume that the potential is zero at infinity.

Short answer

Question: Determine the potential at the surface of a uniformly charged sphere with a charge of \(1.00 \cdot 10^{-6} \mathrm{C}\) and a radius of \(10.0 \mathrm{~cm}\). Answer: The potential at the surface of the sphere is \(8,990 \mathrm{V}\).

Step-by-step solution

Identify the given information

We are given the following information: - Charge of the sphere, \(Q = 1.00 \cdot 10^{-6} \mathrm{C}\) - Radius of the sphere, \(r = 10.0 \mathrm{~cm} = 0.1 \mathrm{~m}\) (converted to meters)

Calculate potential using the formula

We can now calculate the potential at the surface of the sphere using the formula: $$V = \dfrac{kQ}{r}$$ To do so, we will use the value for \(k\) (Coulomb's constant): $$k = 8.99 \cdot 10^{9} \mathrm{\dfrac{N \cdot m^2}{C^2}}$$ Now we can plug in our values: $$V = \dfrac{(8.99 \cdot 10^{9} \mathrm{\dfrac{N \cdot m^2}{C^2}})(1.00 \cdot 10^{-6} \mathrm{C})}{0.1 \mathrm{~m}}$$

Solve for the potential

Now we can solve the equation to find the potential at the surface of the sphere: $$V = \dfrac{(8.99 \cdot 10^{9})(1.00 \cdot 10^{-6})}{0.1}$$ $$V = (8.99 \cdot 10^{3})$$ $$V = 8,990 \mathrm{\dfrac{N \cdot m^2}{C^2}}$$ The potential at the surface of the sphere is \(\mathbf{8,990 \mathrm{V}}\).