Wednesday

April 1, 2015

April 1, 2015

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

- Physics -
**bobpursley**, Sunday, February 7, 2010 at 5:34pmWell, 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 5:47pmSince 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.

**Answer this Question**

**Related Questions**

physics - The electric field at the point x = 5.00cm and y = 0 points in the ...

physics - The electric field at the point and points in the positive direction ...

Physics - The electric field at the point x=5.00 cm and y=0 points in the ...

Physics - 3 There are two charges +Q and –Q/2 that locate at two positions with ...

physics - A uniform electric field exists everywhere in the x, y plane. This ...

Physics - A point charge Q1=-2 microcoulomb is located at x=0, and a point ...

physics - Background pertinent to this problem is available in Interactive ...

Physics - There is a very large horizontal conducting plate in the plane (...

physics - please check: Two point charges, initially 1 cm apart, are moved to a ...

physics - There is a very large horizontal conducting plate in the x-y plane (...