What is the combining ratio of potassium with bromine?
1 answer:
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Answer:
a) 0N/C
b) 2660.7N/C
c) 543.7N/C
Explanation:
To find the electric field in all these cases you take into account the x and y component of the total Electric field for each point.
a)
for P(0,0), only there is the x component of the field because the point is the parallel line that connects both charges:
( 1 )
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
r1: distance to the first charge = 0.150m
r:2 distance to the second charge = 0.150m
Due to in this case the distance r1=r2: you obtain, by replacing in (1):
b)
for P(0.300 , 0 ) you also have only the x component of E, and the electric field generated by each charge are directed toward right:
r1 = 0.300m+0.150m = 0.450m
r2 = 0.300m-0.150m = 0.150m
By replacing r1 and r2 in ( 1 ) you obtain:
c)
for P(0.150 , -0.40) you have both x and y components for E:
the second charge does not contribute for the x component of E.
To find r1 you use Pitagora's theorem:
r2 = 0.40m
the angle is obtain by using a simple trigonometric relation:
Then, by replacing the values of r1, r1, q, theta and k you obtain:
hence, the magnitude of E is 543N/C with an angle of 278° from the positive x axis.