Figure 1. A charged rod of length l and linear charge density λ sits vertically in the y-direction. Point P sits a distance a from the base of the rod.

Sunday, 26 August 2018

1:10 AM

 

 

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Question: 
•p 
a 
Figure 1 _ A charged rod ot length I and linear charge density sits vertically in the y-direction_ Point P sits a distance a from the base ottne rod. 
A rod nas a uniform linear charge density ot A, and a length l. The point P is a distance a to the right of the bottom of the rod as shown in the figure. In this question use Coulomb's constant, 
denoted by k, do not use €0. Take the potential an infinite distance trom a point charge to be 0 (V = 0). 
part 1) 
•p 
a 
Figure 1 _ The rod can be broken into charged infinitesimal slices dy which contribute to the electric field at point P 
Consider a small increment ot the rod with a length dy, a distance y trom the end of the rod. Write an expression to describe the potential due to only this increment at P. 
part 2) 
ay summing the potential trom each ottne increments along the rod derive an expression tor the potential at P due to the rod. 

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part 3) 
using this expression or otnemise come up with an expression for Exp, the electric field in the x direction at P. 
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part 3) 
using this expression or otnemise come up with an expression for Exp, the electric field in the x direction at P. 

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Feedback tor Part 1) 
The charge on the increment q = dy, the distance of the increment trom P is given by d 
a2 + Y2. We can substitute these into the expression for a potential due to a point charge. 
47rE0 r 
dy 
v Pdy = k v/y2+a2 
Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 
Feedback tor Part 2) 
We sum each ottne increments by integrating between y = 0 and y = l. 
v — Ilk 
dy 
v/y24-a2 
dy. 
0 •v/y24-a2 
This is a standard integral so: 
= [In(l+ 12 
- A [In(l+ 12+a2) -In(a)] 
= kA1n( 
Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 

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Feedback tor Part 3) 
The electric field in the direction is given by: 
In order to perform this differentiation we need to know now V changes as changes, in this case distance along the axis is given by a, so near P we can write 
Vp = 12 + — In(x)) 
along the axis perpendicular to the bottom of the rod where x is the distance trom the bottom of the rod. So: 
(In(l + 12 + x2) — In(x)) 
1 
12+x2x(1+ 12+x2) 
12+x2(1+ 12+x2) 
12+x2(1+ 12+x2) 
12+x2) 
12+x2(1+ 12+x2) 
At P, 
= a so: 
12+a2 
Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 
A correct answer is 
, which can be typed in as follows: 
/ sqrt(yA2+aA2) 
A correct answer is k • In 12 + a2 + I — In(a)) • A, which can be typed in as follows: 
A2+aA2 ) 
A correct answer 
, which can be typed in as follows: 
a, 4.a2 
/ (a'sqrt(1A2+aA2) ) 
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Untitled picture.png Machine generated alternative text:
Feedback tor Part 3) 
The electric field in the direction is given by: 
In order to perform this differentiation we need to know now V changes as changes, in this case distance along the axis is given by a, so near P we can write 
Vp = 12 + — In(x)) 
along the axis perpendicular to the bottom of the rod where x is the distance trom the bottom of the rod. So: 
(In(l + 12 + x2) — In(x)) 
1 
12+x2x(1+ 12+x2) 
12+x2(1+ 12+x2) 
12+x2(1+ 12+x2) 
12+x2) 
12+x2(1+ 12+x2) 
At P, 
= a so: 
12+a2 
Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 
A correct answer is 
, which can be typed in as follows: 
/ sqrt(yA2+aA2) 
A correct answer is k • In 12 + a2 + I — In(a)) • A, which can be typed in as follows: 
A2+aA2 ) 
A correct answer 
, which can be typed in as follows: 
a, 4.a2 
/ (a'sqrt(1A2+aA2) )

 

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