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Chapter 1: Vector Analysis Problem 1.1: Using the definitions in Eqs. 1.1 and 1.4, and appropriate diagrams, show that the dot product and cross product are distributive, a) when the three vectors are co-planar. b) in the general case.  Solution   Problem 1.2 Is the cross product associative? $$(\vec{A}\times \vec{B}) \times \vec{C} \overset{?}{=} \vec{A}\times (\vec{B} \times \vec{C})$$ If so, prove it; if not, provide a counterexample (the simpler the better). Solution   Problem 1.3 Find the angle between the body diagonals of a cube.  Solution Problem 1.4 Use the cross product to find the components of the unit vector $\hat{n}$ perpendicular to the shaded plane in Fig. 1.11. Solution   Problem 1.5 Prove the BAC-CAB rule by writing out both sides in component form. Solution Problem 1.6 Prove that $$[\vec{A}\times (\vec{B}\times \vec{C})]+[\vec{B}\times (\vec{C}\times \vec{A})]+[\vec{C}\times (\vec{A}\times \vec{B})] = 0$$ Under what conditions does $\vec{A}...
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Chapter 2 Electrostatics: Problem 2.38

Problem 2.38: A metal sphere of radius $R$, carrying charge $q$, is surrounded by a
thick concentric metal shell (inner radius $a$, outer radius $b$, as in Fig. 2.48). The
shell carries no net charge.
(a) Find the surface charge density $\sigma$ at $R$, at $a$, and at $b$.
(b) Find the potential at the center, using infinity as the reference point.
(c) Now the outer surface is touched to a grounding wire, which drains off charge
and lowers its potential to zero (same as at infinity). How do your answers to
(a) and (b) change?

 


 

 

 

 

 

 

 

 

 

 

Solution:

Create a spherical guassian surface with radius $r$ such that $a\le r\le b$.

 


 

 

 

 

 

 

 

 

 

 

Since electric field inside a conductor is zero. 

Hence the inner part of the shell with inner radius a will get induced with a charge $-q$.

Now a make the shell electrically neutral, positive charge $+q$ will get induced on the outer surface of shell with outer radius $b$. 

 



$a) $ 

$$\sigma_b = \frac{+q}{4\pi b^2}$$ 

$$\sigma_a= \frac{-q}{4\pi a^2}$$ 

$$\sigma_R = \frac{+q}{4\pi R^2}$$

$b)$ Let's denote the center of sphere with C so $$V_{(C)} - V_{\infty} = -\int_{\infty}^0 \vec{E}\cdot \vec{dr}$$ 

$$V_{(C)} - 0 = -\big[\int_{\infty}^b \frac{+q}{4\pi \epsilon_0 r^2} dr + \int_b^a 0dr+\int_a^R \frac{+q}{4\pi \epsilon_0 r^2} dr + \int_R^0 0 dr\big]$$ 

$$V_{(C)} = -\big[\frac{+q}{4\pi \epsilon_0} \big( \frac{-1}{r}\big)_{\infty}^b + \frac{+q}{4\pi \epsilon_0} \big(\frac{-1}{r}\big)_a^R\big]$$

$$V_{(C)} = \frac{+q}{4\pi \epsilon_0} \big[\frac{1}{b} + \frac{1}{R} - \frac{1}{a}\big]$$ 

$c)$ If the outer surface of the shell is connected to ground, So the potential of the outer sphere will turn to 0.

Since charges in the sphere + charges on inner side of shell = $+q+(-q) = 0$

Hence the ground will supply extra $-q$ to the outer side of shell to make its charge 0.

Hence the final charge distribution would be $$\sigma_b = 0 $$ $$\sigma_a= \frac{-q}{4\pi a^2}$$ $$\sigma_R = \frac{+q}{4\pi R^2}$$

 Regarding Potential $$V_{(C)} - V_{\infty} = -\int_{\infty}^0 \vec{E}\cdot \vec{dr}$$ $$V_{(C)} - 0 = -\big[\int_{\infty}^b 0 dr + \int_b^a 0dr+\int_a^R \frac{+q}{4\pi \epsilon_0 r^2} dr + \int_R^0 0 dr\big]$$ 

$$V_{(C)} = -\big[\frac{+q}{4\pi \epsilon_0} \big(\frac{-1}{r}\big)_a^R\big]$$

$$V_{(C)} = \frac{+q}{4\pi \epsilon_0} \big[\frac{1}{R} - \frac{1}{a}\big]$$

 

If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.


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