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Chapter 22: Problem 51

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.

Short Answer

Expert verified
Answer: As the radius of the charged balloon increases and approaches R, the electric field at the point R from the center of the balloon decreases. Once the point R is inside the balloon, the electric field becomes zero.

Step by step solution

01

Write down Gauss's Law

Gauss's Law states that the electric flux through any closed surface is equal to the enclosed charge divided by the vacuum permittivity constant \(\epsilon_0\). Mathematically, it can be written as: $$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0}$$ Here, we'll use this to find the electric field due to the charged balloon.
02

Consider the point outside the balloon (r < R)

At any point outside the balloon, the electric field can be treated as if the entire charge is concentrated at the center of the sphere. Hence, we can consider a Gaussian surface in the shape of a sphere with its center at the center of the balloon and radius r (r < R). Applying Gauss's Law to this Gaussian surface, we have: $$\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$$ Since the electric field and the radial vector have the same direction, the dot product becomes E times the total surface area of the Gaussian surface. $$E\oint dA = E \times 4\pi r^2 = \frac{Q}{\epsilon_0}$$ Solving for E, we get: $$E = \frac{Q}{4\pi \epsilon_0 r^2}$$
03

Consider the point inside the balloon (r > R)

At any point within the balloon, the electric field is zero because there is no charge enclosed within it (as all the charge is on the surface). Therefore, even if the radius of the balloon increases, the electric field inside the balloon will always be zero.
04

Analyze the behavior of the electric field as the radius increases

Initially, when the point where the electric field is being measured is outside the balloon (r < R), the electric field decreases as the radius of the balloon increases since its magnitude is inversely proportional to the square of the radius of the Gaussian surface (which is just r). However, once the point is inside the balloon (r > R), the electric field becomes zero since there is no enclosed charge within the Gaussian surface. In summary, as the radius of the balloon increases and approaches R, the electric field at the point R from the balloon's center decreases and eventually becomes zero when the point is inside the balloon.

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Most popular questions from this chapter

Research suggests that the electric fields in some thunderstorm clouds can be on the order of \(10.0 \mathrm{kN} / \mathrm{C}\). Calculate the magnitude of the electric force acting on a particle with two excess electrons in the presence of a \(10.0-\mathrm{kN} / \mathrm{C}\) field.

A uniform sphere has a radius \(R\) and a total charge \(+Q,\) uniformly distributed throughout its volume. It is surrounded by a thick spherical shell carrying a total charge \(-Q,\) also uniformly distributed, and having an outer radius of \(2 R\). What is the electric field as a function of \(R ?\)

Two infinite, uniformly charged, flat nonconducting surfaces are mutually perpendicular. One of the surfaces has a charge distribution of \(+30.0 \mathrm{pC} / \mathrm{m}^{2}\), and the other has a charge distribution of \(-40.0 \mathrm{pC} / \mathrm{m}^{2}\). What is the magnitude of the electric field at any point not on either surface?

A hollow conducting spherical shell has an inner radius of \(8.00 \mathrm{~cm}\) and an outer radius of \(10.0 \mathrm{~cm} .\) The electric field at the inner surface of the shell, \(E_{\mathrm{i}}\), has a magnitude of \(80.0 \mathrm{~N} / \mathrm{C}\) and points toward the center of the sphere, and the electric field at the outer surface, \(E_{\infty}\) has a magnitude of \(80.0 \mathrm{~N} / \mathrm{C}\) and points away from the center of the sphere (see the figure). Determine the magnitude of the charge on the inner surface and the outer surface of the spherical shell.

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 \(r=R\).
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