In the figure below, the particles have charges q1300 nC and q394255 nC, and dis
ID: 1879389 • Letter: I
Question
In the figure below, the particles have charges q1300 nC and q394255 nC, and distance a 4.5 cm. (Assume that the +x-axis is to the right and the +y-axis is up along the page.) (a) What is the x component of the net electrostatic force on particle 3? (b) What is the y component of the net electrostatic force on particle 3? (a) Did you draw the 3 particles on an x axis? Did you draw the force vectors on the third particle, with the tail of the each vector positioned on the particle? Are the vector orientations correct to give a net force of zero? If the third particle is positive, what are the signs of the other 2 particles to give this vector drawing? What is the relation between the 2 force magnitudes? Did you set up an equation expressing that relation? Did you plug in correctly for the distances? (b) Did you repeat this procedure? Additional Materials Section 21.1 Powers of TenExplanation / Answer
given
q1 = -q2 = 300 nC
q3 = -q4 = 255 nC
a = 4.5 cm
Let F31 is the vertical force acting on the particle 3 due to 1, F32 is the vertical at an angle acting on the particle 3 due to charge 2, and F34 is the horizontal force acting on the particle 3 due to charge 4.
The components of the force F32 are as follows:
The horizontal component is,
F32x = -F32*cos45 = -k*q2*q3*cos45/d2
F32y = -F32 * sin45 = -k*q2*q3*sin45/d2
and
d = sqrt(2)*a so
F32x = - 9*109*(300*10-9)*(255*10-9)*cos45/2*(4.5*10-2)2 = -0.120 N
F32y = - 9*109*(300*10-9)*(255*10-9)*sin45/2*(4.5*10-2)2 = -0.120
F31 = k*q1*q3 /a2
F31 = 9*109*(300*10-9)*(255*10-9)/(4.5*10-2)2 = 0.340 N
F34 = k*q3*q4/a2
F34 = 9*109*255*10-9*255*10-9/(4.5*10-2)2 = 0.289 N
The horizontal component of force acting on the particle 3 is calculated as follows:
F3x = F34 - F32x = 0.289 - 0.120 = 0.169 N
The verticle component of force acting on the particle 3 is calculated as follows:
F3y = -F31 -F32y = -0.340 - 0.120 = 0.46 N