Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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?

Short Answer

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

Step by step solution

01

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 \(8.85 \times10^{-12} \frac{C^2}{N m^2}\).
02

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: \(10 \frac{N m^2}{C} = \frac{Q_{enclosed}}{\epsilon_0}\)
03

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 \(\epsilon_0 = 8.85 \times10^{-12} \frac{C^2}{N m^2}\): \(Q_{enclosed} = 10 \frac{N m^2}{C} \times 8.85 \times10^{-12} \frac{C^2}{N m^2}\)
04

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 \(88.5 \times 10^{-12} C\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two parallel, uniformly charged, infinitely long wires carry opposite charges with a linear charge density \(\lambda=1.00 \mu \mathrm{C} / \mathrm{m}\) and are \(6.00 \mathrm{~cm}\) apart. What is the magnitude and direction of the electric field at a point midway between them and \(40.0 \mathrm{~cm}\) above the plane containing the two wires?

An infinitely long, solid cylinder of radius \(R=9.00 \mathrm{~cm},\) with a uniform charge per unit of volume of \(\rho=6.40 \cdot 10^{-8} \mathrm{C} / \mathrm{m}^{3},\) is centered about the \(y\) -axis. Find the magnitude of the electric field at a radius \(r=4.00 \mathrm{~cm}\) from the center of this cylinder.

A solid nonconducting sphere has a volume charge distribution given by \(\rho(r)=(\beta / r) \sin (\pi r / 2 R) .\) Find the total charge contained in the spherical volume and the electric field in the regions \(rR\). Show that the two expressions for the electric field equal each other at \(r=R\).

A spherical aluminized Mylar balloon carries a charge \(Q\) on its surface. You are measuring the electric field at a distance \(R\) from the balloon's center. The balloon is slowly inflated, and its radius approaches but never reaches R. What happens to the electric field you measure as the balloon increases in radius. Explain.

Two charges, \(+e\) and \(-e,\) are a distance of \(0.68 \mathrm{nm}\) apart in an electric field, \(E,\) that has a magnitude of \(4.4 \mathrm{kN} / \mathrm{C}\) and is directed at an angle of \(45^{\circ}\) with the dipole axis. Calculate the dipole moment and thus the torque on the dipole in the electric field.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free