Question

A Van de Graaff generator has a spherical conductor with a radius of \(25.0 \mathrm{~cm}\). It can produce a maximum electric field of \(2.00 \cdot 10^{6} \mathrm{~V} / \mathrm{m}\). What are the maximum voltage and charge that it can hold?

Short answer

Answer: The maximum charge the Van de Graaff generator can hold is approximately \(1.39 \cdot 10^{-5} \mathrm{~C}\) and the maximum voltage is approximately \(5.00 \cdot 10^4 \mathrm{~V}\).

Step-by-step solution

Recall the electric field near a charged sphere

The formula for the electric field near a charged sphere is given by: $$E = \frac{kQ}{r^2}$$ Where E is the electric field, k is the electrostatic constant \((8.99 × 10^9 Nm^2C^{-2})\), Q is the charge on the sphere, and r is the radius of the sphere. In this case, we know the maximum electric field, E and the radius, r. We need to find the maximum charge, Q that the sphere can hold.

Rearrange the formula to solve for Q

We need to rearrange the electric field formula to solve for the charge Q. The rearranged formula will look like this: $$Q = \frac{E \cdot r^2}{k}$$ Now we can plug in the given values and calculate the maximum charge.

Calculate the maximum charge

Plugging in the values for E, r, and k into the rearranged formula: $$Q = \frac{(2.00 \cdot 10^6 \mathrm{~V/m}) \cdot (0.25 \mathrm{~m})^2}{(8.99 × 10^9 \mathrm{~Nm^2C^{-2}})}$$ After solving this expression, we find that the maximum charge Q is approximately: $$Q \approx 1.39 \cdot 10^{-5} \mathrm{~C}$$

Recall the formula for voltage

Now, we need to find the maximum voltage the generator can hold. The formula for voltage near a charged sphere is given by: $$V = \frac{kQ}{r}$$ Where V is the voltage, k is the electrostatic constant, Q is the charge on the sphere, and r is the radius of the sphere. We know the values of k, r, and we found Q in the previous step. We can plug these values into the formula to find the maximum voltage.

Calculate the maximum voltage

Plugging in the values for k, Q, and r into the voltage formula: $$V = \frac{(8.99 × 10^9 \mathrm{~Nm^2C^{-2}}) \cdot (1.39 \cdot 10^{-5} \mathrm{~C})}{(0.25 \mathrm{~m})}$$ After solving this expression, we find that the maximum voltage V is approximately: $$V \approx 5.00 \cdot 10^4 \mathrm{~V}$$ So, the Van de Graaff generator can hold a maximum charge of \(1.39 \cdot 10^{-5} \mathrm{~C}\) and a maximum voltage of \(5.00 \cdot 10^4 \mathrm{~V}\).

Key concepts

Electric Field

Understanding the electric field is crucial when discussing the Van de Graaff generator. An electric field represents the force per unit charge exerted on a small positive test charge placed at a point in space. It’s a vector field, meaning it has both magnitude and direction, and it’s typically strongest near charged objects and diminishes with distance.

The electric field (\(E\)) around a solitary, spherical charge distribution, like the charged dome of a Van de Graaff generator, can be visualized as lines radiating outward. The density of these lines represents the strength of the electric field at different points. The formula for the electric field near a charged sphere shows how the magnitude of the electric field (\(E\) in volts per meter, V/m) is directly proportional to the charge (\(Q\) in coulombs, C) on the sphere and inversely proportional to the square of the distance (\(r\) in meters, m) from the center of the sphere to the point of interest:
\[\begin{equation}E = \frac{kQ}{r^2}\end{equation}\]
Understand that as the distance from the sphere increases, the electric field strength decreases following the inverse square law, showcasing the intricate balance between electrical force, charge, and distance in electrostatic phenomena.

Electrostatic Constant

The electrostatic constant, represented by the symbol (\(k\)), is a fundamental parameter in the equations governing electrostatics. Also known as Coulomb's constant, it has a value of approximately \(8.99 \times 10^9 Nm^2C^{-2}\) and forms a part of Coulomb's law, which describes the force between two charges.

The electrostatic constant relates the electric force between charges to the product of the charges and inversely to the square of the distance separating them. It’s this constant that ensures the electric force remains consistent across all calculations, acting as a bridge between the abstract mathematical formula and real-world measurements. When computing electric field strength or voltage, as in the Van de Graaff generator exercise, it is this constant that pairs with the charge and distance to yield a value directly applicable to physical situations.

Voltage Calculation

Voltage, also called electric potential difference, is a measure of electric potential energy per unit charge. It’s a scalar quantity, meaning it has magnitude but no direction, and in the case of a charged sphere, it tells us how much work would be required to bring a unit positive charge from infinity to a point near the sphere.

To calculate the voltage (\(V\) in volts, V) at the surface of a sphere, you use the formula:
\[\begin{equation}V = \frac{kQ}{r}\end{equation}\]
This formula reveals that voltage increases directly with the amount of charge on the sphere but decreases as the radius of the sphere grows. In the context of a Van de Graaff generator, the voltage calculation determines the electrical potential that can be built up at the generator's surface, which has critical implications for the generator's ability to store and discharge electrical energy.

Charge on a Sphere

The charge on a spherical conductor, like the dome of the Van de Graaff generator, distributes itself evenly across the surface due to the repulsive force that like charges exert on each other. This phenomenon is also influenced by the shape of the conductor – a sphere ensures that the surface charge distribution remains uniform.

When solving for the charge (\(Q\) in coulombs, C) on the sphere, we derive it from the known electric field and radius by reconfiguring the formula for the electric field. As shown in our solution, charge is the product of the electric field strength and the square of the radius, divided by the electrostatic constant:
\[\begin{equation}Q = \frac{E \times r^2}{k}\end{equation}\]
The sphere's charge influences both the electric field and the voltage it can create. An important aspect of working with charged spheres is ensuring safety, as a large amount of charge can lead to a significant potential difference, which can cause sparks or electrical discharge in the surrounding air – a typical characteristic observed in Van de Graaff generators during demonstrations.