Question

Three charged objects are located at the vertices of a right triangle. Charge A (+5.0 uC) has Cartesian coordinates (0,4); charge B (-5.0 uC) is at the origin; charge C (+4.0 uC) has coordinates (5,0), where coordinates are in metres. What is the net force on charge A?

Answer 1

here,

qA = 5 uC = 5 * 10^-6 C is at (x1,y1) = (0 , 4) m

qB = - 5 uC = - 5 * 10^-6 C is at (x2,y2) = (0 , 0) m

qC = 4 uC = 4 * 10^-6 C is at (x3,y3) = (5 , 0) m

theta = arctan(4/5) = 42.96 degree

the net force on charge A , Fa = K * qA * qB /4^2 *(-j) + K * qA * qC /(4^2 + 5^2) * ( - cos(theta) i + sin(theta) j)

Fa = 9 * 10^9 * 5 * 10^-6 * 10^-6 * (- 0.3125 j - 0.076 i + 0.061 j) N

Fa = ( - 3.42 i - 11.3 j) * 10^-3 N

the magnitude of force on A , |Fa| = sqrt(3.42^2 + 11.3^2) * 10^-3 N = 11.8 N

the direction of force , phi = arctan((-11.3)/(3.24)) = 261.3 degree counterclockwise from +X axis