Here is my problem The only issue I\'m havin is finding the anglein the second p

Question

Here is my problem The only issue I'm havin is finding the anglein the second part. I have this so far: F1= (8.9*109N.m2/C2)(25*10-6C)(25*10-6C)/(0.30m)2 = 61.81 F2= (8.9*109N.m2/C2)(25*10-6C)(25*10-6C)/(0.70m)2 = 11.35 F3= (8.9*109N.m2/C2)(25*10-6C)(25*10-6C)/[(0.70m)2+(0.30 m)2] = 9.59 Angle =tan-1 [W/L]= tan-1[30cm/70cm] = 23.2 The net force is along x -axis is Fx =-F1 - F3 sin = -65.59 The net force is along y -axis is Fy = -F2 - F3 cos = -20.16 Direction is            =tan-1 [Fy/Fx]= tan-1 (-20.16/-65.59) I plugged this into my calculator and got 17.086 but I keepgetting the wrong answer when I put that in. Any clues into whatI'm doing wrong here? Angle =tan-1 [W/L]= tan-1[30cm/70cm] = 23.2 The net force is along x -axis is Fx =-F1 - F3 sin = -65.59 The net force is along y -axis is Fy = -F2 - F3 cos = -20.16 Direction is            =tan-1 [Fy/Fx]= tan-1 (-20.16/-65.59) I plugged this into my calculator and got 17.086 but I keepgetting the wrong answer when I put that in. Any clues into whatI'm doing wrong here?

Explanation / Answer

Given :   Charge ( q ) =+25.0 µC L = 70.0 cm W = 30.0cm. k = 9 * 10-9N.m2/C2 as we know that    F = k q2 /r2 F1= ( 9*109N.m2/C2)(25*10-6C)(25*10-6C)/(0.30m)2 = 62.5 F2= ( 9*109N.m2/C2)(25*10-6C)(25*10-6C)/(0.70m)2 = 11.5 F3= (8.9*109N.m2/C2)(25*10-6C)(25*10-6C)/[(0.70m)2+(0.30 m)2] = 9.7 Angle =tan-1 [W/L]            =tan-1[0.30m / 0.70m]            =23.2 The net force is along x -axis is Fx =-F1 - F3 sin     = -66.5 The net force is along y -axis is Fy = -F2 - F3 cos      = -20.3 Direction is            =tan-1 [Fy/Fx]            = -----    Solve it I hope it helps you Angle =tan-1 [W/L]            =tan-1[0.30m / 0.70m]            =23.2 The net force is along x -axis is Fx =-F1 - F3 sin     = -66.5 The net force is along y -axis is Fy = -F2 - F3 cos      = -20.3 Direction is            =tan-1 [Fy/Fx]            = -----    Solve it I hope it helps you

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