For equal charges are placed on the corners of one face of a cube.?

Question

Four equal charges of +3.6×10-6 C are placed on the corners of one face of a cube of edge length 8.0 cm. A charge of -3.6×10-6 C is placed at the center of the cube.

What is the magnitude of the force on the charge at the center of the cube?

civil_av8r · Answer

F_net = q*E_net
E_net = the vector sum of all the electric fields.

Distance from each + charge to the center
r = sqrt (4^2 + 4^2 + 4^2) cm = 6.9 cm = 0.069 m

Here's the easy rationalization trick to this problem... if the 4 charges where place on a x/y coordinate plane, all the x/y components of the electric field cancel (equal and opposite x/y components).  However, since this is a 3D problem, the z-components all add together.

E_net = 4*k*q/r^2*cos (45) = 27.2 * 10^6 N/C

45 is the angle the vectors makes with the z-axis.

F_net = q*E_net = 98 N... direction is not important, but because of the negative point charge placed in this electric field, the force vector points in the opposite direction as the electric field vector.

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For equal charges are placed on the corners of one face of a cube.?

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