**
Field Theory - Online Tutorial 1 - New **

**by Kevin Tang ,
new update May 26, 2014**

A charge ρ_{l
}[C/m] is uniformly distributed on the line −a < z < a. and
point P given P (0, b, 0)

Find the following...

* Please review this question, as I didn't solve it fully today ~ good luck :)

* Notice: any
**BOLD** letter means that variable is a VECTOR.

1a) Draw the diagram

1b) Find **r**,
**r'**, **r **− **r'**, | **r **− **r'** | and dQ.

1c) **E **at
point (0, b, 0)

**1a)** Draw the diagram..

Solving Time: 3 min

__SOLUTION 1a)__

Shown in Figure #1 below...

*Figure # 1*

**1b)** Find **r**, **r'**, **r **− **r'**, | **r **−
**r'** | and dQ.

Solving Time : 5 min

Let's have fun and give a try...(click the answer that you think it is correct)

Which one is **r**?

A | B | C | D |

Which one is **r'**?

A |
B |
C |

Than you can get can get **r **− **r'** and | **r **− **r'** |

Next, which one is dQ?

A |
B |
C |

__SOLUTION 1b) __

Answer for **r**, **r'**, **r
**− **r'**, | **r **− **r'** | and dQ.

Detail solution shown as following (shown in Figure #2)...

*Figure # 2*

**1c)** **E **at point (0, b, 0)

Solving Time : 15 min

__SOLUTION 1c) __

Actually the solution is much longer than I expected, please take a good look...

First, sub everything into d**E**

*Figure # 3*

Second, split the integral into 2 PART.

Remember, as I said in class, you can visually see that **a _{z}
**part (part 2) will cancel out to zero
due to symmetric,

however, here I am still going to show you how its lead to zero.

Now, we start with **a _{y}** part (part 1) of the
integral...(shown in Figure #4)

you need to applied integral by trigonometric substitution (or you can directly check it on Integral Table)

*Figure # 4*

.

.

.

.

Part 2, **a _{z}** part of integral, its will cancel out and
lead to zero as shown following (shown in Figure #5)...

*Figure # 5*

Last step, sub everything into E, and you will get the final answer (shown in Figure #6) :)

*Figure # 6*

You may leave me any message if you have any questions, I will try to answer ASAP, thanks and good luck :)