Question

In: Physics

A charge of -2.75 µC is fixed at the center of a compass. Two additional charges...

A charge of -2.75 µC is fixed at the center of a compass. Two additional charges are fixed on the circle of the compass, which has a radius of 0.092 m. The charges on the circle are -4.3 µC at the position due north and +5.25 µC at the position due east. What are the magnitude and direction of the net electrostatic force acting on the charge at the center? Specify the direction relative to due east.

Expert Solution

Given that

A charge of -2.75 µC is fixed at the center of a compass

Two additional charges are fixed on the circle of the compass, which has a radius of 0.092 m.

The charges on the circle are -4.3 µC at the position due north and +5.25 µC at the position due east

Then the force between the two charges is given by

F =kq1q2/r2

The charge from the due north is negative. So, it is charge in the middle.

Therefore the force will be repulsive and it directs in south

F1 =9*10⁹*(2.75*10^-6)((4.3*10^-6)/(0.092)² = 106.425*10^-3 /0.008464=12.573N south

Form the other charge one is positive and the other is negative, then force will be towards east

F2 =9*10⁹*(2.75*10^-6)((5.25*10^-6)/(0.092)² = 129.937*10^-3 /0.008464=15.351N east

then the net force is given by

F =Sqrt(F12+F22) =Sqrt[(12.573)2+(15.351)2 =Sqr[158.080+235.653]=Sqrt[393.733]=19.842N

Then the direction is given by

Tan theta =F1/F2 =12.573/15.351 = 0.819

Theta =tan-1(0.819)=0.686 degrees south of east

genius_generous answered 1 year ago

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