Physics
posted by Katherine on .
The electric field at the point x=5.00 cm and y=0 points in the positive x direction with a magnitude of 10.0 N/C. At the point x=10.0 cm and y=0, the electric field points in the positive x direction with a magnitude of 15.0 N/C. Assuming the field is produced by a single point charge, find (a) it's location and (b) the sign and magnitude of its charge

Well, E = kq/r^2
E(5)=10
E(10)=15
E(5)=10=kQ/(p+.05)^2
E(10)=15=kQ/(p+.1)^2
divide the first equation by the second:
2/3 = [(P+.1)/(P=.05)]^2
take the sqrt of both sides, solve for P
Then put P into either equation, solve for Q 
Since the Efield is pointed along the +x axis at locationa along the x axis, the point charge that causes the field but also be on the x axis. The distance from the point charge to (5,0) must be sqrt (2/3) = 0.8165 times the distance from the point to (10,0), because the field is 2/3 as strong at (5,0) as it is at (10,0).
The only way you can make this happen is to have the point charge be positive and located to the left x = 0.