الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

Electric Field
Example 3.6
What is the electric field in the lower left corner of the square as shown
in figure 3.11? Assume that q = 1×10 -7 C and a = 5cm.
Solution
First we assign number to the charges (1, 2, 3, 4) and then determine the
direction of the electric field at the point p due to the charges.
E
E
1
2
1
=
4πε ο
1
=
4πε ο
q
a
2
q
2a
2
+q 1
+q
2
E
3
1 2q
=
4πε a2
ο
Evaluate the value of E 1 , E 2 , & E 3
E 1 = 3.6×10 5 N/C,
E 2x
P E 3
E 2y
E 2
E 1
Figure 3.11
3
-2q
E 2 = 1.8 × 10 5 N/C,
E 3 = 7.2 × 10 5 N/C
Since the resultant electric field is the vector additions of all the fields i.e.
r
E p
r
= E
r
r
1+
E2
+ E3
We find the vector E 2 need analysis to two components
E 2x = E 2 cos45
E 2y = E 2 sin45
E x = E 3 - E 2 cos45 = 7.2×10 5 - 1.8 × 10 5 cos45 = 6 × 10 5 N/C
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