Question

Two equally charged, 3.699 g spheres are placed with 3.592 cm between their centers. When released, each begins to accelerate at 297.727 m/s2. What is the magnitude of the charge on each sphere? Express your answer in microCoulombs.

Answer 1

Given:

The mass of the first sphere is: m1 = 3.699 g.

The mass of the second sphere is: m2 = 3.699 g

The distance between their centers is: d = 3.592 cm

The acceleration of each sphere is: a = 297.727 m/s^2

To find:

Since the spheres are identical in their masses, the force on each sphere is:

`F=ma`

Substitute the values in the above equation and simplify it, we get:

`\begin{gathered} F=3.699\text{ g}*297.727\text{ m/s}^2 \\ \\ F=3.699\text{ g }*\frac{1\text{ kg}}{1000\text{ g}}*297.727\text{ m/s}^2 \\ \\ F=3.699*10^(-3)\text{ kg}*297.727\text{ m/s}^2 \\ \\ F=1.1012\text{ N} \end{gathered}`

This is the force experienced by each sphere and is has a magnitude equal to the magnitude of the electrostatic force.

The electrostatic force of attraction or repulsion between two charges is given by:

`F=(1)/(4\pi\epsilon_0)(q^2)/(r^2)`

Substitute the values in the above equation and simplify it, we get:

`\begin{gathered} 1.1012\text{ N}=\frac{9*10^9\text{ N}\cdot m^2\text{/C}^2* q^2}{(3.592\text{ cm})^2} \\ \\ 1.1012\text{ N}=\frac{9*10^9\text{ N}\cdot m^2\text{ / C}^2* q^2}{(3.592\text{ cm}*\frac{1\text{ m}}{100\text{ cm}})^2} \\ \\ 1.1012\text{ N}=\frac{9*10^9\text{ N}\cdot m^2\text{ /C}^2* q^2}{(3.592*10^(-2))^2\text{ m}^2} \\ \\ 1.1012\text{ N}=\frac{9*10^9\text{ N.m}^2\text{/C}^2* q^2}{1.2902*10^(-3)\text{ m}^2\text{ }} \\ \\ 1.1012\text{ N}=6.9757*10^(12)\text{ N/C}^2* q^2 \\ \\ \end{gathered}`

Rearranging the above equation and simplify it, we get:

`\begin{gathered} q^2=\frac{1.1012\text{ N}}{6.9757*10^(12)\text{ N/C}^2} \\ \\ q=\sqrt{1.5786*10^(-13)\text{ C}^2} \\ \\ q=0.3973*10^(-6)\text{ C} \\ \\ q=0.3973\text{ }\mu\text{C} \end{gathered}`

Final answer:

The magnitude of the charge on each sphere is 0.3973 microcolumns.