Question
Asked Sep 4, 2019

ME: I found the promblem on the web site (book:university-physics-with-modern-physics-14th-edition, isbn 9780321973610, chapter 21, question 27.87)

only the charges are different, however I got in to touble with the conclusion.  within the Eq give(Part A), the denomiator it equals to zero; at least with my math.

Thanks for the help

-Av

Given:

Two 1.20m nonconducting wires meet at a right angle. One segment carries 2.00 μC of charge distributed uniformly along its length, and the other carries − 2.00 μC distributed uniformly along it, as shown in the figure

Part A:

Find the magnitude of the electric field these wires produce at point P, which is 60.0 cm from each wire

Part B:

Find the direction of the electric field these wires produce at point P, which is 60.0 cm from each wire.(Suppose that the y-axis directed vertically.)
 
Part C:
If an electron is released at P, what is the magnitude of the net force that these wires exert on it?
 
 
Part D
If an electron is released at P, what is the direction of the net force that these wires exert on it?(Suppose that the y-axis directed vertically.)
 
 
 

 

 

-1.20 m_
+ + + + + + + + +
P
120 m
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-1.20 m_ + + + + + + + + + P 120 m

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Expert Answer

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Step 1

Consider the figure,

1.20 m
+ + + + +
d 0.6m
E
1.20 m
VE
net
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1.20 m + + + + + d 0.6m E 1.20 m VE net

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Step 2

(a) The magnitude of filed due to a line of charge is.

is the line charge density
a is the half of the length of the line
E
d is the distance from P to line
2TEO dda
L
put ,and a=1
2
L
1 Q
1
4,7TE dd +(L/2)
=9x103 Nm2/c2 v 2 x10C
0.6m
1
X
V(0.6m)
+(1.20 m/2)
- 35355 N/C
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is the line charge density a is the half of the length of the line E d is the distance from P to line 2TEO dda L put ,and a=1 2 L 1 Q 1 4,7TE dd +(L/2) =9x103 Nm2/c2 v 2 x10C 0.6m 1 X V(0.6m) +(1.20 m/2) - 35355 N/C

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Step 3

The magnitude of electric field electric field due to positive and negative charge  are same but electric field due to negatively charged wir...

E E_E
net
= -35355N/ci-35355 N/cj
E2 E
+
net
(-35355N/ci)+(-35355N/ci)
5.00 x 104 N/C
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E E_E net = -35355N/ci-35355 N/cj E2 E + net (-35355N/ci)+(-35355N/ci) 5.00 x 104 N/C

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