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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}\times (\vec{B}\times \v

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, 

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

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


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