Recent content by veevee

Oh ok, I see. It's 0. For some reason i thought that the +Q and the -Q would cancel out.

A conductor carries a net charge +Q. There is a hollow cavity inside the conductor that contains a point charge -Q. In electrostatic equilibrium,what is the charge on the outer surface of the conductor? A. -2Q B. -Q C. 0 D. Q E. 2Q I know that the answer is 0, however, I cannot...

hmm... so since the length from the q charge to the Q charge is inversely proportional to the force felt by Q (1/y), the closer Q is to both q's, the more force it experiences. But if it was the lined up with them on the x axis, the net force would be 0. So, do I need to make it as close to...

Suppose 2 drops with equal charge q are on the x-axis at x=+-a. Find the maximum electric force felt by a third drop with charge Q at an arbitrary point on the y axis. from the symmetry we can deduce that the x components cancel and the net force is in the y direction. the y component is...

the problem was i was entering 1 sig fig not 2 for Fy lol...online homework submissions...they'll waste your time for the simplest reasons..thanks though :)))

r13=sqrt(x2+y2)=sqrt(4+1)=2.2m r23=2m x=2 y=1

(x,y) axis in meters q1=60microC (0,1) q2=-34microC (2,0) q3=13microC (2,2) Fx3=? Fy3=? i used superposition F(net on 3)=F13+F23 broke it down into x and y components Fx=(kq1q3/r13^2)(x/r13) Fy=(kq1q3/r13^2)(y/r13) - (kq2q3/r23^2) plugging in the numbers i got Fx=1.3...