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Charge Sharing among Three Metal Balls Four identical conducting balls are attached to insulating supports that sit on the floor as shown below. One ball has charge \(Q\); its support is fixed in space. The other three balls are uncharged but their supports can be moved around. Describe a procedure (that involves only moving and/or bringing balls into contact) that will leave the \(+Q\) ball with its full charge and give the three originally uncharged balls charges \(q,-q / 2\), and \(-q / 2\). You may assume that \(q

Short Answer

Expert verified
Achieve the charges \(+Q\), \(+q\), \(-\frac{q}{2}\), and \(-\frac{q}{2}\) by contact and separation.

Step by step solution

01

Initial Setup

We start with four conducting balls: one has a charge of \(+Q\) and the other three are uncharged. The charged ball is fixed in position, while the other three can be moved around.
02

Contact the Charged Ball with an Uncharged Ball

Bring one of the uncharged balls into contact with the charged ball. The charge \(+Q\) will equally distribute between them, so each ball will now have \(+\frac{Q}{2}\) charge.
03

Separate the Balls

Separate the balls from Step 2. Now, one of the previously uncharged balls has \(+\frac{Q}{2}\) charge while the originally charged ball remains with \(+\frac{Q}{2}\) charge.
04

Repeat with a Second Uncharged Ball

Next, bring the second uncharged ball into contact with the ball that has \(+\frac{Q}{2}\) charge. The total charge \(+\frac{Q}{2}\) will distribute evenly between these two balls, giving each \(+\frac{Q}{4}\) charge.
05

Separate and Final Contact

Separate these balls. Now, we have two balls each with \(+\frac{Q}{4}\) charge. Bring this new \(+\frac{Q}{4}\) charged ball into contact with the final uncharged ball. They will split the charge, resulting in each ball having \(+\frac{Q}{8}\) charge.
06

Calculate Final Charges

Now compute the charges on all balls: - Initially, one ball with \(+Q\).- After Step 2, one with \(+\frac{Q}{2}\) and one with \(+\frac{Q}{2}\).- After Step 4, one with \(+\frac{Q}{4}\), one with \(+\frac{Q}{4}\).- After Step 5, one with \(+\frac{Q}{8}\), one with \(+\frac{Q}{8}\).According to the problem constraints, the charges need to be labeled \(q, -q/2, -q/2\). Therefore, these can be labeled appropriately across the uncharged balls to satisfy the \(+q\) and \(-q/2\) requirement while maintaining the original charge ball unaffected.

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Key Concepts

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

Charge Distribution
When conductive objects come into contact, charge distribution occurs. Imagine two objects. If one is charged and the other is not, electrons move to distribute the charge between them. This is especially important with conductive balls because they are made of material that allows electrons to move freely.

When you place two conductive balls in contact, any charge present on one ball can move onto the other. If a charged ball with charge \(+Q\) meets an uncharged one, the charge will redistribute equally, assuming both are the same size and material. This is why, after contact, each ball will end up with half the total initial charge of the charged ball. Thus, starting with a charge \(+Q\), each ball would then have \(+\frac{Q}{2}\) after contact.
  • Conductive objects allow free movement of charge.
  • Contact results in equal distribution of the charge if objects are identical.
  • This principle guides the steps necessary to distribute charge accurately between balls.
Conducting Balls
Conducting balls are ideal for demonstrating electrostatic principles because they allow electrons to move unimpeded across their surfaces. Their spherical shape ensures an even distribution without concentrated spots of charge.
  • Spheres distribute charge evenly over their surface when they are conductive.
  • Contact between conducting spheres allows charge to redistribute for equilibrium.
This property makes them perfect for experiments involving charge conservation and distribution. It means that in any setup, the total charge before and after contact should remain constant, redistributed among the spheres they contact.
Insulating Supports
Insulating supports are essential as they prevent the transfer of charge from the conducting balls to their environment or each other unintentionally. By holding the balls with non-conductive materials, we ensure that charge transfers occur only during intended contact, such as when handling the spheres directly for redistribution.

Without insulation, charges could leak, making it difficult to control the charge distribution accurately. For example, an insulating rod used to hold a charged sphere ensures that no charge escapes to the ground or another object, maintaining the integrity of the experiment.
Charge Conservation
Charge conservation is a fundamental principle in electrostatics. It states that the total charge in an isolated system remains constant, though the charge can be transferred between objects within the system. When conducting balls come into contact and separate, they must adhere to this rule.
  • The sum of charges before contact equals the sum after separation.
  • This principle helps predict charge distribution on balls after contact.
For example, in our exercise, when the charged ball is dispersed over multiple contacts, the principle of charge conservation ensures that the initial charge \(+Q\) is entirely shared among the four balls throughout the process without any charge being lost. Each step involving contact and separation shows this principle clearly, with specific targets for the final charge quantity on each ball.

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Most popular questions from this chapter

Charge Induction by a Dipole A point dipole \(\mathbf{p}\) is placed at \(\mathbf{r}=\mathbf{r}_{0}\) outside a grounded conducting sphere of radius \(R\). Use Green's reciprocity (and a comparison system with zero volume charge density) to find the charge drawn up from ground onto the sphere.

The Force between Conducting Hemispheres (a) A spherical metal shell is charged to an electrostatic potential \(V\). Cut this shell in half and pull the halves infinitesimally apart. Find the force with which one hemisphere of the shell repels the other hemisphere. (b) A spherical capacitor is formed from concentric metal shells with charges \(\pm Q\) and radii \(b>a\). Cut this capacitor in half and pull the halves infinitesimally apart. Find the force with which the two halves repel one another.

Concentric Cylindrical Shells A capacitor is formed from three very long, concentric, conducting, cylindrical shells with radii \(a

Bounds on Parallel-Plate Capacitance Let \(C\) be the capacitance of capacitor formed from two identical, flat conductor plates separated by a distance \(d\). The plates have area \(A\) and arbitrary shape. When \(d \ll \sqrt{A}\), we know that the capacitance approaches the value \(C_{0}=A \epsilon_{0} / d\). (a) If \(\delta \mathbf{E}=\mathbf{E}-\mathbf{E}_{0}\), prove the identity $$ \int_{V} d^{3} r|\mathbf{E}|^{2}-\int_{V} d^{3} r\left|\mathbf{E}_{0}\right|^{2}=\int_{V} d^{3} r|\delta \mathbf{E}|^{2}+2 \int_{V} d^{3} r \mathbf{E}_{0} \cdot \delta \mathbf{E} $$ (b) Let \(\mathbf{E}=-\nabla \varphi\) be the actual field between the finite- area plates and let \(\mathbf{E}_{0}=-\nabla \varphi_{0}\) be the uniform field that would be present if \(A\) were infinite. Use the identity in part (a) to prove that \(C>C 0\), using \(V\) as the volume between the finite-area plates. Assume that the potentials \(\varphi\) and \(\varphi_{0}\) take the same (constant) values on the plates.

Concentric Spherical Shells Three concentric spherical metallic shells with radii \(c>b>a\) have charges \(e_{c}, e_{b}\), and \(e_{a}\), respectively. Find the change in potential of the outermost shell when the innermost shell is grounded.

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