Answer to Question #200957 in Field Theory for Mallory

Question

 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.

Answer 1

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

$$E=\dfrac{V}{d}=\dfrac{416\ \dfrac{J}{C}}{0.10\ m}=4160\ \dfrac{N}{C}.$$

(b) Let's wtite the conditions of the equilibrium in projections on axis $x$- and $y$:

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

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

$$sin\theta=\dfrac{x}{L},$$ $$\theta=sin^{-1}(\dfrac{x}{L}),$$ $$\theta=sin^{-1}(\dfrac{0.011\ m}{1.0\ m})=0.63^{\circ}.$$

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

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

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

$$F_e=Tsin\theta=4.6\cdot10^{-3}\ N\cdot sin0.63^{\circ}=5.1\cdot10^{-5}\ N.$$

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

$$F_e=qE,$$ $$q=\dfrac{F_e}{E}=\dfrac{5.1\cdot10^{-5}\ N}{4160\ \dfrac{N}{C}}=1.23\cdot10^{-8}\ C.$$