Two identical plastic balls of mass \(10 \mathrm{gm}\) each are hung by threads with length \(30 \mathrm{~cm}\) from a common point, as shown below. The balls are each charged with the same charge \(q\) and repel each other until they come to rest with a horizontal separation of \(30 \mathrm{~cm}\). a) Sketch the electric field produced by the two balls. b) Draw the force vectors on the right-hand ball. c) What is the charge \(q\) in each ball?

Since both balls have the same charge, they will repel each other, resulting in an electric field that points away from each ball. The electric field lines will start from one ball, curve, and terminate at the other ball. It will be symmetrical due to their equal mass and charge.

There are three force vectors acting on the right-hand ball: 1. Weight: A downward vector due to gravitational force (\(F_g = mg\)), where \(m\) is the mass of the ball and \(g\) is the acceleration due to gravity. 2. Tension: The force exerted by the thread, acting along the thread's length and pointing towards the common point where the threads are attached. 3. Electric force: A horizontal force pointing leftwards, acting due to repulsion between the charged balls (\(F_e\)). These three forces should balance each other, meaning their vectorial sum should be equal to zero.

As the balls are at rest, we can conclude that their net force is zero. The electric force acts horizontally, and the tension and gravitational force can be decomposed into their vertical and horizontal components. Let's denote the angle between the threads and the vertical as \(\theta\). Then, the horizontal component of the tension force, \(T_x = T\sin{\theta}\), balances the electric force. And the vertical component, \(T_y = T\cos{\theta}\), balances the gravitational force. Now we can find the tangent of the angle \(\theta\): \(\tan{\theta} = \frac{T_x}{T_y} = \frac{15}{15} = 1\) \(\theta = 45^{\circ}\) Since the electric force and the horizontal component of the tension force are equal, we can write: \(F_e = T\sin{\theta}\)

Coulomb's law defines the repulsive force between two identical charges as: \(F_e = k\frac{q^2}{r^2}\), where \(k\) is the electrostatic constant, \(q\) is the charge, and \(r\) is the distance between the charges. We can now substitute the expression for \(F_e\) from the previous step: \(k\frac{q^2}{r^2} = T\sin{\theta}\) Solving for the charge \(q\), we get: \(q = \sqrt{\frac{T\sin{\theta}r^2}{k}}\) We know the gravitational force is equal to the vertical component of the tension force: \(mg = T\cos{\theta}\) Solving for the tension force \(T\), we get: \(T = \frac{mg}{\cos{\theta}}\) Now we can substitute the value of \(T\) and plug in the given values: \(q = \sqrt{\frac{(10^{-2} \text{ kg})(9.8 \text{ m/s}^2)}{\cos{45^{\circ}}}\frac{(0.3 \text{ m})^2}{9 \times 10^9 \text{ Nm}^2/\text{C}^2}\sin{45^{\circ}}}\) \(q \approx 7.94 \times 10^{-8} \text{ C}\) Hence, the charge on each ball is approximately \(7.94 \times 10^{-8} \text{ C}\).