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Chapter 4: Q60P (page 717)

Two charges are placed on the x-axis: one, of 2.50 μC, at the origin and the other, of -3.50 μC, at x = 0.600 m (Fig. P21.60). Find the position on the x-axis where the net force on a small charge +q would be zero.

Short Answer

Expert verified

The net force on charge +q would be zero when +-q is at distance 3.27 m at negative x-axis.

Step by step solution

01

Position of Charges

The unknown charge is positive like charge q1,therefore the position of the charge +q must be left to charge q1, as shown below.

02

Net Force

The net force that equals zero could be given by

Fnet=F1+F20=14πεoq1|+q|d2+14πεoq2|+q|(d+0.600m)2q1d2=q2(d+0.600m)2dd+0.600m=q1q2

03

Substitution

Put values to find d:

dd+0.600m=|2.50μC|∣−3.50μCd=3.27m

Therefore, the net force on charge +q would be zero when +-q is at distance 3.27 m at negative x-axis

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