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Chapter 1: Problem 80

Force in a soap bubble ** Like the charged rubber balloon described at the end of Section 1.14, a charged soap bubble experiences an outward electrical force on every bit of its surface. Given the total charge \(Q\) on a bubble of radius \(R\), what is the magnitude of the resultant force tending to pull any hemispherical half of the bubble away from the other half? (Should this force divided by \(2 \pi R\) exceed the surface tension of the soap film, interesting behavior might be expected!)

Short Answer

Expert verified
The force tending to separate the hemispheres of the bubble is \( F = \frac{KQ^2}{4 \pi R^2} \).

Step by step solution

01

Understand the problem

A soap bubble is charged with a total charge \(Q\) and has a radius \(R\). The goal is to calculate the force exerted by one half of the bubble on the other half due to the distribution of charges.
02

Calculate the electric field

We know that the electric field \( E \) due to a uniformly charged sphere at points outside the sphere is given by, \( E = \frac{KQ}{r^2} \), where \( K = 8.99 \times 10^9 N \cdot m^2/C^2 \) is Coulomb's constant, \( Q \) is the charge on the sphere and \( r \) is the distance from the center of the sphere.
03

Calculate the force

The electric field is due to all the charges on the bubble, so the force on a charge \( dq \) due to the electric field is given by \( dF = E \cdot dq \) = \( \frac{KQdq}{r^2} \), where \( dq \) is the charge on an infinitesimally small area of the hemisphere.
04

Integrate to find total force

We integrate \( dF \) over the surface of the hemisphere to get the total force. The total charge \( Q \) is uniformly distributed over the surface area of the sphere, so \( dq = Q \cdot \frac{da}{4 \pi R^2} \), where \( da \) is an infinitesimally small area of the hemisphere. Substituting for \( dq \) we get \( dF = \frac{KQ^2da}{4 \pi R^4} \). Integrating over the hemisphere, \( F = \int dF = \frac{KQ^2}{4 \pi R^4} \int da \). For a hemisphere, the area \( A = 2 \pi R^2 \), so \( F = \frac{KQ^2}{4 \pi R^2} \).
05

Finalise the answer

So, the force tending to pull any hemispherical half of the bubble away from the other half is \( F = \frac{KQ^2}{4 \pi R^2} \).

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Key Concepts

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

Electric Field
The electric field is a fundamental concept in physics that represents the presence and magnitude of force per unit charge around a charged object. In essence, it's an invisible field that surrounds electric charges and exerts force on other electric charges within the field. When considering a charged soap bubble, as in our exercise, the electric field describes how the charges distributed across the bubble's surface will influence each other.

For a soap bubble with a total charge of \(Q\), the electric field at a point outside the bubble (such as on the opposing side of the bubble surface) is given by the expression \(E = \frac{KQ}{r^2}\), where \(K\) is Coulomb's constant, and \(r\) is the distance from the center of the bubble to the point of interest. This field causes the similar charges to repel each other, creating a force that can affect the bubble's stability.
Surface Tension in Physics
Surface tension is a physical phenomenon that occurs at the boundary between two mediums, such as the air and the liquid forming a soap bubble. It's created by the cohesive forces between liquid molecules, which are stronger for molecules within the liquid compared to those on the surface due to the imbalance of forces.

These surface molecules pull toward the rest of the liquid, creating a 'skin' that makes the bubble hold its shape. Surface tension is the reason soap bubbles are spherical, as the shape minimizes surface area for a given volume, thus minimizing energy due to surface tension. In our exercise, we compare the electrostatic force trying to pull the hemispheres of a charged soap bubble apart with the force of surface tension holding it together. If the electrostatic force exceeds the force of surface tension, the bubble may deform or even burst.
Coulomb's Law
\( Coulomb's Law \) describes the force of attraction or repulsion between two stationary charged particles. It states that the magnitude of the electrostatic force \( F \) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance \( r \) between them, expressed as \( F = K\frac{|q_1 q_2|}{r^2} \), where \( K \) is Coulomb's constant.

In the context of the charged soap bubble problem, Coulomb's Law helps us understand the force of repulsion between the two halves of the bubble due to their charges. By integrating this principle over the hemisphere of the bubble, we can find the resultant force that might challenge the bubble's surface integrity - bridging the gap between individual particle interactions and macroscopic physical effects.

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Most popular questions from this chapter

Field from two charges ** A charge \(2 q\) is at the origin, and a charge \(-q\) is at \(x=a\) on the \(x\) axis. (a) Find the point on the \(x\) axis where the electric field is zero. (b) Consider the vertical line passing through the charge \(-q\), that is, the line given by \(x=a\). Locate, at least approximately, a point on this line where the electric field is parallel to the \(x\) axis.

Field from two sheets : Two infinite plane sheets of surface charge, with densities \(3 \sigma_{0}\) and \(-2 \sigma_{0}\), are located a distance \(\ell\) apart, parallel to one another. Discuss the electric field of this system. Now suppose the two planes, instead of being parallel, intersect at right angles. Show what the field is like in each of the four regions into which space is thereby divided.

Gravity vs. electricity (a) In the domain of elementary particles, a natural unit of mass is the mass of a nucleon, that is, a proton or a neutron, the basic massive building blocks of ordinary matter. Given the nucleon mass as \(1.67 \cdot 10^{-27} \mathrm{~kg}\) and the gravitational constant G as \(6.67 \cdot 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \mathrm{s}^{2}\right)\), compare the gravitational attraction of two protons with their electrostatic repulsion. This shows why we call gravitation a very weak force. (b) The distance between the two protons in the helium nucleus could be at one instant as much as \(10^{-15} \mathrm{~m}\). How large is the force of electrical repulsion between two protons at that distance? Express it in newtons, and in pounds. Even stronger is the nuclear force that acts between any pair of hadrons (including neutrons and protons) when they are that close together.

Hole in a plane : (a) A hole of radius \(R\) is cut out from a very large flat sheet with uniform charge density \(\sigma\). Let \(L\) be the line perpendicular to the sheet, passing through the center of the hole. What is the electric field at a point on \(L\), a distance \(z\) from the center of the hole? Hint: Consider the plane to consist of many concentric rings. (b) If a charge \(-q\) with mass \(m\) is released from rest on \(L\), very close to the center of the hole, show that it undergoes oscillatory motion, and find the frequency \(\omega\) of these oscillations. What is \(\omega\) if \(m=1 \mathrm{~g},-q=-10^{-8} \mathrm{C}, \sigma=10^{-6} \mathrm{C} / \mathrm{m}^{2}\), and \(R=0.1 \mathrm{~m} ?\) (c) If a charge \(-q\) with mass \(m\) is released from rest on \(L\), a distance \(z\) from the sheet, what is its speed when it passes through the center of the hole? What does your answer reduce to for large \(z\) (or, equivalently, small \(R\) )?

Uniform field strength * We know from the example in Section \(1.11\) that the electric field inside a solid sphere with uniform charge density is proportional to \(r\). Assume instead that the charge density is not uniform, but depends only on \(r\). What should this dependence be so that the magnitude of the field at points inside the sphere is independent of \(r\) (except right at the center, where it isn't well defined)? What, should the dependence be in the analogous case where we have a cylinder instead of a sphere?
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