Find the net electrostatic force on the 2.4 mu c charge. Three point-charges lie

Question

Find the net electrostatic force on the 2.4 mu c charge. Three point-charges lie along a straight line. Charge q_1 = 4.0 mu C and is located at 4.0 m from charge q_2 = 2.0 mu C. Where along the line must the third charge q_3 = -2.0 muC be placed so that the net force on it is zero Four point-charges are located at the comers of a square of side l as shown in the figure. If q_1 = -q_2 = e, q_3 = q_4 = -2e, and l = 25.0cm. what is the resultant elextrostatie force on q_2 Consider the configuration of point charges shown in the figure. Charge q_1 = -2.0 mu C and charge q_2 = 2.0 mu C .What are the unit vectors pointing from charges q_1 and q_2 to point P Suppose l place a charge Q = -3.0 mu C at point P. What force will it feel Remember, force is a vector.

Explanation / Answer

18) let distance from q1 to third charge be x, and distance from third charge to second be d - x
q1 = 4*10^-6 C
q2 = 2*10^-6 C
q3 = -2*10^-6 C
d = 4.0 m

Solve kq1q3/x² = kq2q3/(d - x)² ===> sqrt(q2/q1) = (d-x)/x
(4-x)/x = 0.707 =====> x= 2.34 m
q3 needs to be at 2.34 m

19) Force due to q1 on q2 = F12= kq1q2/l^2 = (9*10^9*1.6*10^-19*1.6*10^-19)/0.25^2

F12 = 3.69*10^-27 N (along left or -x-direction)   

Force due to q4 on q2 = F42= kq1q2/l^2 = (9*10^9*1.6*10^-19*2*1.6*10^-19)/0.25^2

F42 = 7.37*10^-27 N (along down or -y-direction)

Force due to q3 on q2 = F32= kq1q2/(2l^2) = (9*10^9*1.6*10^-19*2*1.6*10^-19)/(2*0.25^2)

F32 = 3.69*10^-27 N (at 45 degree)

So, F32x = 3.69*10^-27 *sin45 = 2.61*10^-27 N (along right or x-direction)

  F32y = 3.69*10^-27 *cos45 = 2.61*10^-27 N (along up or y-direction)

Now force on q2 along x-direction Fx= (2.61-3.69)*10^-27 = 1.08*10^-27 N (along left or -x-direction)

force on q2 along y-direction Fy= (2.61-7.37)*10^-27 = 4.76*10^-27 N (along down or -y-direction)

Net force = F= 4.88*10^-27 N (at 77.21 degree)

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