Question

The net electric field in a region of space is a vector sum of two parts. The first has a magnitude of E1 = 1100 N/C and points at an angle of 25° from the positive x axis towards the positive y axis. The second has a magnitude of E2 = 1600 N/C and points 50° from the positive y axis towards the negative x axis. A 5.0 kg mass, that is charged to 0.10 C, is moving at 10 m/s in the positive x direction when it enters the field. What is the magnitude and direction of the acceleration of the mass when it enters the field. (ignore gravitational effects)

Answer 1

E_netx = E_1x+E_2x = E₁cos25 + E₂sin(-50) = 1100cos25-1600sin(50) = -228.73 N/C

E_nety = E_1y+E_2y = E₁sin25 + E₂sin(-50) = 1100sin25+1600cos(50) = 149.33 N/C

F_netx = q*E_netx = (0.10C)*(-228.73 N/C) = -22.87 N

F_nety = q*E_nety = (0.10C)*(1493.34 N/C) = 149.33 N

a_netx = F_netx/m = (-22.87 N)/(5.0kg) = - 4.53 m/s²

a_nety = F_nety/m = (149.33 N)/(5.0kg) = 29.87 m/s²

magnitude = a_net =sqrt[a_netx² + a_nety² ]=sqrt(4.53^2+29.87^2) = 30.21 m/s²

direction = θ=tan^-1(a_nety/a_netx)= tan^-1(29.87/4.53) = 81.38 …above negative X axis.