Symettry makes the problem go away.
On the adjacent corners: F=kqq/s^2 * cos45
note that you resolve the force into two components, one component in the direction of the diagonal (above), and a perpendicular. However, the perpendicular to the normals from each corner cancel each other, as they are opposite directions. So you are left with the forces from corners in the direction of the normal
Fcorners=2*kqq/s^2 *cos45
now add the force from the diagonal charge.
Fdiag=kqq/(s*sqrt2)^2=1/2 kqq/s^2
add that to Fcorners, and you have it.
Make certain you understand this problem, you will see it again.
Pls help me out.
Four positive and equal charges are located dt the corners of a square of side 20cm.if each charge has a magnitude 2microcoulumb,calculate the force acting on the charge at the lower left corner of the square.
1 answer
1 answer