Question

Two small metal cubes with masses 2.0 g and 4.0 g are tied together by a 5.0-cm-long massless string and are at rest on a frictionless surface. Each is charged to +2.0 mC. a. What is the energy of this system? b. What is the tension in the string? c. The string is cut. What is the speed of each cube when they are far apart?

Short answer

The value of the energy of the system is$0.72J$

The value of tension in the string is $14.4N$

The value of the speed of each cube are $10.95m/s$and $22m/s$

Step-by-step solution

Definition of  the potential energy and Tension

The electric potential of the two charges is $U=\frac{k{q}_1{q}_2}{r}$

and force is $F=\frac{k{q}_1{q}_2}{r^2}$

$$\begin{gathered} U=\frac{k{q}_1{q}_2}{r} \\ r=0.05cm \\ k=8.99\times {10}^9/N\times m^2 \\ {q}_1{q}_2=2.0\times {10}^{6}C \end{gathered}$$

Applying all data

$U=\frac{\left(8.99\times {10}^9/N\times m^2\right)\left(2.0\times {10}^{6}C\right)}{0.05cm}=0.72J$

role="math" localid="1648131453480" width="159" height="172">F=kq1q2r2r=0.05cmk=8.99×109/N×m2q1q2=2.0×106C

Applying all data

$F=\frac{\left(8.99\times {10}^9/N\times m^2\right)\left(2.0\times {10}^{6}C\right)}{0.05c{m}^2}=14.4N$

Definition of  the energy of cube system 

$$\begin{gathered} m_1=2\times {10}^{-3}Kg \\ {m}_2=4\times {10}^{-3}Kg \end{gathered}$$

As the K.E is equals to localid="1648131983690" $\frac{1}{2}m{v_1}^2+\frac{1}{2}m{v_2}^2$
Applying all data
localid="1648131933744" $$\begin{gathered} v_2=10.95m/s \\ {v}_1=22m/s \end{gathered}$$