Problem 78 - University Physics with Modern Physics

Question

Suppose you have a large spherical balloon and you are able to measure the component \(E_{n}\) of the electric field normal to its surface. If you sum \(E_{n} d A\) over the whole surface area of the balloon and obtain a magnitude of \(10 \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}\) what is the electric charge enclosed by the balloon?

Step-by-Step Solution

Answer

Answer: The electric charge enclosed by the balloon is \(88.5 \times 10^{-12} C\).

Step 1: Write down Gauss's Law.

First, let's recall Gauss's Law, which states that the electric flux through a closed surface is equal to the total enclosed electric charge divided by the vacuum permittivity. Mathematically, it can be written as: \(\oint E_n dA = \frac{Q_{enclosed}}{\epsilon_0}\) Where \(\oint E_n dA\) is the total electric flux through the surface, \(Q_{enclosed}\) is the enclosed electric charge, and \(\epsilon_0\) is the vacuum permittivity which is equal to \(8.85 \times10^{-12} \frac{C^2}{N m^2}\).

Step 2: Use the given information.

We are given that the surface integral of \(E_n dA\) over the entire surface area of the balloon is \(10 \frac{N m^2}{C}\). Thus, we can write our equation as: \(10 \frac{N m^2}{C} = \frac{Q_{enclosed}}{\epsilon_0}\)

Step 3: Solve for the enclosed electric charge.

Now, we have to find \(Q_{enclosed}\). Rearranging the equation, we get: \(Q_{enclosed} = 10 \frac{N m^2}{C} \times \epsilon_0\) Plugging in the value for the vacuum permittivity \(\epsilon_0 = 8.85 \times10^{-12} \frac{C^2}{N m^2}\): \(Q_{enclosed} = 10 \frac{N m^2}{C} \times 8.85 \times10^{-12} \frac{C^2}{N m^2}\)

Step 4: Calculate the enclosed electric charge.

Now we can calculate the value for \(Q_{enclosed}\): \(Q_{enclosed} = 88.5 \times10^{-12} C\) The electric charge enclosed by the balloon is \(88.5 \times 10^{-12} C\).

Key Concepts

Electric Flux

Electric flux is a measure of the electric field passing through a given area. It's graphically represented by the number of electric field lines penetrating a surface. For our balloon scenario, imagine an invisible net catching the lines of the electric field generated by the charge inside. The total electric flux through our balloon's surface is expressed mathematically as the integral of the normal component of the electric field, denoted as , multiplied by the differential area element, dA, over the entire surface. In simple terms, you are summing up the electric field's strength at every tiny piece of the balloon's surface area. The outcome of 10 Nm²/C in our exercise indicates the total electric flux and directly relates to the charge inside under Gauss’s Law.

Enclosed Electric Charge

The enclosed electric charge refers to the total amount of charge residing within a closed surface. It is the source of electric fields that we detect outside the surface. In the context of Gauss's Law, it's the charge that creates the electric flux we have just discussed. To figure out the charge inside our hypothetical balloon, you must know the total electric flux passing through its surface. It's like knowing the amount of water flowing out from a sponge, and consequently, estimating how much water the sponge holds. In our mathematical journey, the step-by-step solution uses this principle to determine the balloon's hidden treasure – the amount of charge inside it.

Vacuum Permittivity

Vacuum permittivity, denoted as _0, is a fundamental physical constant. It represents how much electric field is 'permitted' in the vacuum. Think of it as the measure of the vacuum's willingness to allow electric field lines to pass through itself. In terms of its role in Gauss's Law, it's a scaling factor that relates the electric flux through a surface to the enclosed charge. In essence, it ensures that the units balance out correctly, and we can calculate the enclosed charge based on the electric flux we observed. The value 8.85 × 10^-12 C²/Nm² is the accepted standard in physics and is crucial for the last steps of the solution where we determine the enclosed charge.

Integral Calculus in Physics

Integral calculus plays an essential role in physics, and Gauss's Law is a prime example of its application. It allows us to sum infinitesimally small pieces of a quantity to determine a total. In our scenario, we integrate or sum up the tiny contributions of the electric field over the balloon’s surface to obtain the total electric flux. It's like calculating the total amount of paint needed for a wall by adding up the paint required for each tiny section. The integral symbol ∮ represents this 'summing up' process over a closed surface, and it is the key to finding the solution to our problem. Integral calculus, thus, turns a conceptually challenging problem into a solvable puzzle using mathematical tools.