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Field from two shells One of two nonconducting spherical shells of radius \(a\) carries a charge \(Q\) uniformly distributed over its surface, the other carries a charge \(-Q\), also uniformly distributed. The spheres are brought together until they touch. What does the electric field look like, both outside and inside the shells? How much work is needed to move them far apart?

Short Answer

Expert verified
The electric field is zero everywhere inside the spheres and \(E = 2k \frac{Q}{r^2}\) outside, pointing away from the centre of the positive charge and towards the centre of the negative one. The work required to move the spheres far apart is \(2Q^2k/a\).

Step by step solution

01

Determine the electric field due to each sphere individually

Consider the spheres separately. As stated by Gauss' law, the electric field inside a sphere of uniform charge is zero, and outside it is \(E = k \frac{Q}{r^2}\), where \(k\) is Coulomb's constant and \(r\) is the distance from the centre of the sphere. For the positive sphere the electric field points away from the centre, while for the negative sphere it points towards the centre.
02

Superpose the electric fields

When the spheres are brought together, their electric fields superpose. Inside the overlapping volume of the spheres where both fields exist, they are equal in magnitude but opposite in direction, so they cancel each other. Therefore, the electric field everywhere inside the spheres is zero. In the region outside, the electric fields are in the same direction and add up, leading to an electric field of magnitude \(E = 2k \frac{Q}{r^2}\), which points away from the centre of the positive sphere and towards the centre of the negative one.
03

Determine the work done to move the spheres apart

The work required to separate the spheres will correspond to the potential energy change associated with moving a charge \(Q\) from the surface of the negative sphere to infinity. Since the electric field outside the shells is the sum of the fields due to each shell alone, this is the same as the work required to move a charge from a distance \(a\) (radius of each shell) to infinity in an electric field of \(E = 2k \frac{Q}{r^2}\). Therefore, the required work is equal to \(W = \int_{a}^{\infty} F dr = \int_{a}^{\infty} Q E dr = 2Q^2k \int_{a}^{\infty} dr / r^2 = 2Q^2k/a\).

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Key Concepts

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

Gauss's Law
Gauss's Law is a fundamental principle in electromagnetism that relates the electric field and electric charge distributions. It states that the electric flux through a closed surface is proportional to the charge enclosed within that surface. Mathematically, it is expressed as: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]where \( \vec{E} \) is the electric field, \( d\vec{A} \) is a differential area, \( Q_{\text{enc}} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space. When applying Gauss's Law to problems involving spherical symmetry, like the spherical shells in this exercise, it simplifies calculations because of the uniform distribution of charge. - Inside a uniformly charged spherical shell, the electric field is zero due to symmetry, and outside it follows the inverse square law. This means that outside the shell, the electric field decreases with the square of the distance from the center.
Spherical Shells
Spherical shells, especially when non-conducting, have intriguing characteristics due to their symmetry in charge distribution. When we deal with such shells having charge uniformly distributed over their surfaces, these shells exhibit certain predictable behaviors.- Inside the shell, as per Gauss's Law, the electric field is zero. This occurs because the symmetrically spread-out charge results in an overall cancellation of electric field vectors when calculated at any internal point.- Outside the shell, the electric field corresponds to that of a point charge placed at the center of the sphere. Let's consider our original exercise where we have two shells brought together with charges \( Q \) and \(-Q \). The electric fields they produce are crucial in understanding the electric field behavior both inside and outside the combined shells when superposing their fields. The fields within the touching shells cancel each other, resulting in zero net field internally, while externally, the fields from each shell add up, resulting in an amplified field.
Coulomb's Constant
Coulomb's Constant, denoted as \( k \), is a key factor in the mathematical description of electric forces and fields. It is a fundamental constant with a value approximately equal to \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \). This constant appears in Coulomb's Law, which describes the force between two point charges. The formula is:\[ F = k \frac{|q_1 q_2|}{r^2} \]where \( F \) is the force between the charges, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, \( r \) is the distance between them, and \( k \) is Coulomb's Constant. In the context of spherical shells, Coulomb's Constant helps determine the magnitude of the electric field outside the shells when they are brought together. Specifically, it's crucial in calculating the work needed to separate two charged shells, as we've seen in the exercise where the work was determined using the integral formula that involves \( k \). This constant allows us to transform theoretical concepts into real-world applications where electric forces come into play.

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Most popular questions from this chapter

Field on the earth A sphere the size of the earth has \(1 \mathrm{C}\) of charge distributed evenly over its surface. What is the electric field strength just outside the surface? What is the potential of the sphere, with zero potential at infinity?

Curl of a gradient The electric field equals the negative gradient of the potential, that is, \(\mathbf{E}=-\nabla \phi\). Show that this implies that the curl of \(\mathbf{E}\), which we can write as \(\nabla \times \mathbf{E}\), is identically zero. Do this by: (a) calculating \(\nabla \times \nabla \phi\) in Cartesian coordinates; (b) making judicious use of Stokes' theorem.

Finding the potential The following vector function represents a possible electrostatic field: $$ E_{x}=6 x y, \quad E_{y}=3 x^{2}-3 y^{2}, \quad E_{z}=0 $$ (We have ignored a multiplicative factor with units of \(\mathrm{V} / \mathrm{m}^{3} \mathrm{nec}-\) essary to make the units correct.) Calculate the line integral of \(\mathbf{E}\) from the point \((0,0,0)\) to the point \(\left(x_{1}, y_{1}, 0\right)\) along the path that runs straight from \((0,0,0)\) to \(\left(x_{1}, 0,0\right)\) and thence to \(\left(x_{1}, y_{1}, 0\right)\). Make a similar calculation for the path that runs along the other two sides of the rectangle, via the point \(\left(0, y_{1}, 0\right)\). You ought to get the same answer if the assertion above is true. Now you have the potential function \(\phi(x, y, z)\). Take the gradient of this function and see that you get back the components of the given field.

Dividing the charge We have two metal spheres, of radii \(R_{1}\) and \(R_{2}\), quite far apart from one another compared with these radii. Given a total amount of charge \(Q\) which we have to divide between the spheres, how should it be divided so as to make the potential energy of the resulting charge distribution as small as possible? To answer this, first calculate the potential energy of the system for an arbitrary division of the charge, \(q\) on one sphere and \(Q-q\) on the other. Then minimize the energy as a function of \(q\). You may assume that any charge put on one of these spheres distributes itself uniformly over the surface of the sphere, the other sphere being far enough away so that its influence can be neglected. When you have found the optimum division of the charge, show that with that division the potential difference between the two spheres is zero. (Hence they could be connected by a wire, and there would still be no redistribution. This is a special example of a very general principle we shall meet in Chapter 3: on a conductor, charge distributes itself so as to minimize the total potential energy of the system.)

Interstellar dust * An interstellar dust grain, roughly spherical with a radius of \(3 \cdot 10^{-7} \mathrm{~m}\), has acquired a negative charge such that its potential is \(-0.15\) volt. How many extra electrons has it picked up? What is the strength of the electric field at its surface?

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