Posted by **Katherine** on Sunday, February 7, 2010 at 5:22pm.

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

- Physics -
**bobpursley**, Sunday, February 7, 2010 at 5:34pm
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

- Physics -
**drwls**, Sunday, February 7, 2010 at 5:47pm
Since the E-field 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.

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