Ch 21 Electric Charge and Field

Learning Objectives

  1. Understand the principles of the inverse square law of electrostatic force to conductors and insulators
  2. Understand and work with electric forces and fields
  3. Calculate electric fields for continuous distributions of charge

Quiz

1 Charge Q is uniformly distributed along a thin ring of radius aa. (a) Derive an expression for the distance xx along the axis of the ring where the electric field EE is a maxinum. (b) what charge placed at this location (point P) would make E=0E = 0 at the center of the ring? Explain your answer.
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Solution

2 A small 80g insulating sphere is suspended from poing P by a silk thread that is 50cm long. The sphere bears an unknow electric charge Q. A positive point charge q=+2.0μCq = +2.0 \mu C is brought to a position directly below P, and the sphere is repelled to a new position, 30cm to the right of qq, as shown. The charge QQ is closest to?
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Solution

(1) The electric force F=kqQr2F = \dfrac{kqQ}{r^2}.
(2) The weight of the insulating sphere is W=mgW = mg
(3) WF=tan⁡θ=43\dfrac{W}{F} = \tan\th = \dfrac{4}{3}
Combine (1), (2) and (3), we have

Q=3mgr24kqm=8×10−2kgq=2×10−6Cr=0.3m⇒Q=3×8×10−2×10×0.324×9×109×2×10−6=3×10−6C\begin{aligned} Q &= \frac{3mgr^2}{4kq}\\ m &= 8 \times 10^{-2} kg\\ q &= 2\times 10^{-6} C \\ r &= 0.3m\\ \To Q &= \frac{3 \times 8 \times 10^{-2} \times 10 \times 0.3^2}{4 \times 9 \times 10^{9} \times 2 \times 10^{-6}}\\ &= 3 \times 10^{-6}C \end{aligned}