Question

If a Van de Graaff generator has an electric potential of \(1.00 \cdot 10^{5} \mathrm{~V}\) and a diameter of \(20.0 \mathrm{~cm},\) find how many more protons than electrons are on its surface.

Short answer

Based on the given information, find the number of excess protons on the surface of a Van de Graaff generator with an electric potential of \(1.00 \times 10^5 V\) and a diameter of \(20.0 cm\).

Step-by-step solution

Write down the formula to find the charge.

We can use the formula for electric potential on the surface of a sphere, which is given by: $$V = \frac{kQ}{r}$$ where \(V\) is the electric potential, \(k\) is the electrostatic constant (approximately \(8.99 \times 10^9 Nm^2/C^2\)), \(Q\) is the total charge stored on the surface, and \(r\) is the radius of the sphere.

Calculate the total charge stored on the surface.

By rearranging the given formula, we can find the charge stored on the surface of the Van de Graaff generator: $$Q = \frac{V \cdot r}{k}$$ We have the values of electric potential \(V = 1.00 \times 10^5 V\) and diameter of the generator \(d = 20.0 cm\). First, convert the diameter into the radius (in meters) by dividing by 2 and convert centimeters to meters: \(r = 0.1 m\). Substitute these values into the equation: $$Q = \frac{(1.00 \times 10^5 V)(0.1 m)}{8.99 \times 10^9 Nm^2/C^2}$$ Now, calculate \(Q\).

Calculate the number of excess protons on the surface.

The elementary charge value is \(e = 1.60 \times 10^{-19} C\). To find the number of excess protons, divide the total charge by the elementary charge: $$n = \frac{Q}{e}$$ Substitute the calculated value of \(Q\) from Step 2 and divide by the elementary charge value to find the number of excess protons.

Final answer.

Calculate the value of \(n\) using the values from the previous steps and state the number of excess protons on the surface of the Van de Graaff generator.