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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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