A metal sphere with radius ra is supported on an insulating...

A metal sphere with radius $r_{a}$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_{b} .$ There is charge $+q$ on the inner sphere and charge $-q$ on the outer spherical shell. (a) Calculate the potential $V(r)$ for (i) $r < r_{a} ;$ (ii) $r_{a} < r < r_{b} ;$ (iii) $ r> r_{b}$ . Hint: The net potential is the sum of the potentials due to the individual spheres.) Take $V$ to be zero when $r$ is infinite. (b) Show that the potential of the inner sphere with respect to the outer is $$V_{a b}=\frac{q}{4 \pi \epsilon_{0}}\left(\frac{1}{r_{a}}-\frac{1}{r_{b}}\right)$$ (c) Use Eq. $(23.23)$ and the result from part (a) to show that the electric field at any point between the spheres has magnitude $$E(r)=\frac{V_{a b}}{\left(1 / r_{a}-1 / r_{b}\right)} \frac{1}{r^{2}}$$ (d) Use Eq. $(23.23)$ and the result from part (a) to find the electric field at a point outside the larger sphere at a distance $r$ from the center, where $r>r_{b}$ . (e) Suppose the charge on the outer sphere is not $-q$ but a negative charge of different magnitude, say $-Q$ . Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different.

A metal sphere with radius $r_{a}$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_{b} .$ There is charge $+q$ on the inner sphere and charge $-q$ on the outer spherical shell. (a) Calculate the potential $V(r)$ for (i) $r < r_{a} ;$ (ii) $r_{a} < r < r_{b} ;$ (iii) $ r> r_{b}$ . Hint: The net potential is the sum of the potentials due to the individual spheres.) Take $V$ to be zero when $r$ is infinite. (b) Show that the potential of the inner sphere with respect to the outer is
$$V_{a b}=\frac{q}{4 \pi \epsilon_{0}}\left(\frac{1}{r_{a}}-\frac{1}{r_{b}}\right)$$
(c) Use Eq. $(23.23)$ and the result from part (a) to show that the electric field at any point between the spheres has magnitude
$$E(r)=\frac{V_{a b}}{\left(1 / r_{a}-1 / r_{b}\right)} \frac{1}{r^{2}}$$
(d) Use Eq. $(23.23)$ and the result from part (a) to find the electric field at a point outside the larger sphere at a distance $r$ from the center, where $r>r_{b}$ . (e) Suppose the charge on the outer sphere is not $-q$ but a negative charge of different magnitude, say $-Q$ . Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different.

Key Concepts

Spherical Symmetry

Spherical symmetry in a charge distribution implies that the properties of the system, such as the electric field and potential, depend only on the radial distance from the center and not on the angular coordinates. This symmetry significantly simplifies the analysis and mathematical computations of electrostatic problems by reducing them to one-dimensional integrals or algebraic forms.

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. It is a powerful tool in determining electric fields for highly symmetric charge distributions such as spheres, where it simplifies the computation of both the electric field and, through integration, the potential. The law is pivotal when considering the electric fields inside and outside conductors.

Conductors in Electrostatics

Conductors exhibit unique behavior in electrostatic conditions where free charges reside on the surface and the interior remains field-free in equilibrium. The potential throughout a conductor is constant, and boundary conditions at the conductor's surface play a critical role in solving electrostatic problems. This concept is essential when analyzing systems involving metallic objects and shells.

Electrostatic Potential

This concept refers to the work done per unit charge in bringing a test charge from infinity to a specified point in an electric field. It is a scalar quantity that simplifies the analysis of electric fields, particularly in problems involving spherical symmetry. Calculating the potential often involves integrating the electric field or summing contributions from individual charge elements using the principle of superposition.

Superposition Principle

In electrostatics, the superposition principle states that the total potential or electric field created by multiple sources is the algebraic sum of the potentials or fields due to each individual source. This principle is especially useful when dealing with complex systems that can be decomposed into simpler symmetric charge distributions, allowing for straightforward calculation of the overall potential.

Transcript

00:01 This is the problem based for calculation of a charge body.

00:08 Here it is given solid sphere conducting of radius ra on the insulating strength and conducting shell of radius rb plus q is charge on inner conductor and minus q is charge on outer conductor.

00:47 In the first part we have to find potential as a function of r first r is less than r a second r is greater than r a but less than r v third r is greater than r v b part show that potential difference between the spare a and v to b q upon four pi as a epsilon not 1 upon r a minus 1 upon r b r b c part using our results so that electric field is given by potential of a b 1 upon r a minus 1 upon r b into 1 upon r square d part electric field at r is greater than r v e if minus q is replaced by minus q then prove that result of part b and c not affected but d will change let us start solving it starting with part a potential is given by kq by rv this is the answer of part a b part this is 4 r is less than r b for r is greater than r a but less than r b it will be k q by r minus k q by r v that is k q one upon r minus one upon r b c part for r is k q greater than r b potential will be 0 because k q by r so it becomes zero these are the parts of a first i have to make a correction here this is second and this is third b part potential of a 1 upon 4 pi epsilon not and potential of b is 0 so potential of b is 0 so potential of ab will be q upon 4 pi epsilon not c part electric field is defined as negative of potential gradient that is minus tail over del r q upon 4 pi epsilon not 1 upon r minus 1 upon r b so you will get q upon 4 pi epsilon not r squared now potential of ab is q upon 4 pi epsilon not 1 upon r a minus 1 upon r b so from here q can be written as 4 pi a b a b upon 1 upon r a substituting in equation 1 you will get v a b upon 1 upon r a b upon r a minus 1 upon r b into 1 upon r square...

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