Hey how can I solve this? The drawing shows an equilateral triangle, each side o
ID: 1669766 • Letter: H
Question
Hey how can I solve this? The drawing shows an equilateral triangle, each side of whichhas a length of 4.32 cm. Point charges are fixed to each corner, asshown. The 4.00 C charge experiences a net force due to thecharges qA and qB. This netforce points vertically downward and has a magnitude of 722 N.Determine a) charge qA,b) charge qB. Thanks! Hey how can I solve this? The drawing shows an equilateral triangle, each side of whichhas a length of 4.32 cm. Point charges are fixed to each corner, asshown. The 4.00 C charge experiences a net force due to thecharges qA and qB. This netforce points vertically downward and has a magnitude of 722 N.Determine a) charge qA,b) charge qB. Thanks!Explanation / Answer
The unknown charges can be determined using Coulomb’slaw. Since the net force points downwards, we know that the unknowncharges are both negative. Since the net force points downwards, we know that the unknowncharges have the same magnitudes. The horizontal components of the forces cancel (one pointing to theright and the other to the left). The vertical components of the forces add (reinforce) to providethe observed net force. F=k{[(4.00*10^-6C)|qA|]/r^2}=k{[(4.00*10^-6C)|qB|/r^2} The magnitudes of the vertical components of the forces from qA andqB are: F cos 30 degrees F=k{[(4.00*10^-6C)|qA|]/r^2}cos30+k{[(4.00*10^-6C)|qB|/r^2}cos30 F=2k{[(4.00*10^-6C)|qA|]/r^2}cos30 and solving for the magnitude of the charge: |qA|=(F)r^2/2k(4.00*10^-6C)cos30 work it out from here.. whatever answer you get for qA will be thesame for qB