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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 1 Vector Analysis: Problem 2.10

Problem 2.10: A charge $q$ sits at the back corner of a cube, as shown in Fig. 2.17.
What is the flux of E through the shaded side?

 


 

 

 

 

 

 

 

 

Solution: 

Surround the charge by 8 such cubes such that 4 cubes are above and 4 cubes are below and all are geometrically equivalent.


 

 

 

 

 

 

 

 

 

Consider, the bigger cube having all the 8 small cubes.

By guass law, the total flux would be $$\oint \vec{E}\cdot \vec{da} = {Q_{enc}\over \epsilon} = {q\over \epsilon}$$ 

Let $\phi$ be the flux through the required surface (The dark one in the diagram). 

There would be $4\times 6 $ such faces and all are geometrically equivalent. Hence, the flux through them would be same and summation of them would be the total flux. Hence, 

$$\oint \vec{E}\cdot \vec{da} = {24 \phi} = {q\over \epsilon_0}$$ 

$$\therefore \phi = {q\over 24\epsilon_0}$$  


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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