Question

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

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

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.

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: $$E = \frac{Q}{4\pi \epsilon_0 r^2}$$

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.

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.

Key concepts

Electric Field

The electric field is a fundamental concept that describes the region around a charged particle or object where a force would be exerted on other charges. Imagine it like an invisible field of influence, extending out from the charged object.

The strength and direction of this field are represented by electric field lines.

• The lines go away from positive charges and toward negative charges.
• The density of these lines indicates how strong the electric field is.

For a spherical charge distribution, like our balloon, the electric field at a distance from the center can be calculated using Gauss's Law. If the point of interest is outside the charge distribution, the entire charge can be considered as if located at the sphere's center. This leads to an electric field given by \(E = \frac{Q}{4\pi \epsilon_0 r^2}\), where \(Q\) is the total charge, and \(r\) is the distance from the center.

Inside a conductor that is spherical and has all its charge on the surface, like the balloon, the electric field inside is zero. This happens because any internal point doesn't have any net enclosed charge affecting it.

Spherical Balloon

A spherical balloon can be seen as a round object with a uniform charge distribution on its surface.

• As it inflates, the physical dimensions of the balloon change, but the total charge remains constant.
• Due to its shape, the balloon fits perfectly into Gauss’s Law calculations.

The symmetry of a spherical shape simplifies the application of Gauss's Law, helping us assume uniform electric fields and easily calculate the effects on a point placed at a distance from the center.

The interesting aspect of this spherical balloon setup is how we analyze changes in electric field intensity. As the balloon inflates, any point close to the surface eventually ends up inside, where no electric field is present. This demonstrates the very localized nature of electric fields when dealing with surfaces, fundamentally different from infinite or very large charge distributions.

Charge Distribution

Charge distribution refers to how electric charge is spread over a particular area or volume. For the spherical balloon scenario, the charge \(Q\) is uniformly distributed over its surface.

• Uniform distribution means that every point on the balloon’s surface has the same charge density.
• As the balloon inflates, the total charge remains the same, but its distribution changes because the surface area increases.

Even if the total charge remains constant, the effective influence (electric field) at a distance changes because of this varying distribution. This is crucial when considering Gauss's Law, as the law directly relates the electric field to the enclosed charge.

This concept of charge distribution helps us understand why the electric field inside the balloon becomes zero. As the charge spreads over a larger inflated surface, any point inside the balloon ends up not enclosed by any charge, leading to no net electric field influence.