الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

Lectures in General Physics
E y = -E 1 - E 2 sin45 = -3.6×10 5
- 1.8 × 10 5 sin45 = - 4.8 × 10 5 N/C
E E x + E y
2 2
= = 7.7 × 10 5 N/C
E
E
y
θ = tan −1
= - 38.6 o
x
Example 3.7
In figure 3.12 shown, locate the point at which the electric field is zero?
Assume a = 50cm
Solution
-5q 2q
V S E P
1 2
1
- +
a
d
a+d
E 2
Figure 3.12
To locate the points at which the electric field is zero (E=0), we shall try all
the possibilities, assume the points S, V, P and find the direction of E 1 and
E 2 at each point due to the charges q 1 and q 2 .
The resultant electric field is zero only when E 1 and E 2 are equal in
magnitude and opposite in direction.
At the point S E 1 in the same direction of E 2 therefore E cannot be zero in
between the two charges.
Dr. Hazem Falah Sakeek