Answer to Question #200957 in Field Theory for Mallory

Question #200957

 A ping pong ball of mass 4.686E-4 kg, hangs from a light thread 1.0 m long between two vertical parallel plates 10.0 cm apart. When the potential difference across the plates is 416.0 J/C, the ball comes to equilibrium 1.1 cm to one side of its original position. (a) Calculate the electric field strength between the plates. (b) Calculate the tension in the thread (c) Calculate the magnitude of the electric force acting on the ball (c) Calculate the charge on the ball. Note, please neglect any vertical deflection of the ball.




1
Expert's answer
2021-05-31T08:39:20-0400

(a) By the definition of the electric field strength, we have:


E=Vd=416 JC0.10 m=4160 NC.E=\dfrac{V}{d}=\dfrac{416\ \dfrac{J}{C}}{0.10\ m}=4160\ \dfrac{N}{C}.

(b) Let's wtite the conditions of the equilibrium in projections on axis xx- and yy:


Tcosθmg=0,(1)Tcos\theta-mg=0, (1)TsinθFe=0.(2)Tsin\theta-F_e=0. (2)

We can find the angle θ\theta from the geometry:


sinθ=xL,sin\theta=\dfrac{x}{L},θ=sin1(xL),\theta=sin^{-1}(\dfrac{x}{L}),θ=sin1(0.011 m1.0 m)=0.63.\theta=sin^{-1}(\dfrac{0.011\ m}{1.0\ m})=0.63^{\circ}.

Finally, we can find the tension in the thread from the first equation:


T=mgcosθ=4.686104 kg9.8 ms2cos0.63=4.6103 N.T=\dfrac{mg}{cos\theta}=\dfrac{4.686\cdot10^{-4}\ kg\cdot9.8\ \dfrac{m}{s^2}}{cos0.63^{\circ}}=4.6\cdot10^{-3}\ N.

(c) We can find the magnitude of the electric force acting on the ball from the second equation:


Fe=Tsinθ=4.6103 Nsin0.63=5.1105 N.F_e=Tsin\theta=4.6\cdot10^{-3}\ N\cdot sin0.63^{\circ}=5.1\cdot10^{-5}\ N.

(d) We can find the charge on the ball as follows:


Fe=qE,F_e=qE,q=FeE=5.1105 N4160 NC=1.23108 C.q=\dfrac{F_e}{E}=\dfrac{5.1\cdot10^{-5}\ N}{4160\ \dfrac{N}{C}}=1.23\cdot10^{-8}\ C.

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