Question

(a) Calculate the potential energy of a system of two small spheres, one carrying a charge of 2.00C and the other a charge of -3.50$\mathit{\mu }$C, with their centers separated by a distance of 0.180 m. Assume that U = 0 when the charges are infinitely separated. (b) Suppose that one sphere is held in place; the other sphere, with a mass of 1.50 g, is shot away from it. What minimum initial speed would the moving sphere need to escape completely from the attraction of the fixed sphere? (To escape, the moving sphere would have to reach a velocity of zero when it is infinitely far from the fixed sphere.)

Short answer

a) The potential energy is $U=-0.252]$

b) Initial speed is $V_a=18.3m/s$

Step-by-step solution

Step-by-Step Solution

Given data:

$\begin{array}{l}{Q}_1=-3.50c\\ Q_2=2.0\mu C\\ d=0.25m\\ m=1.5g\end{array}$

Step 2:

So, the potential between two-point charges is

$\begin{array}{l}U=\frac{k{Q}_1{Q}_{}}{d}\\ =\frac{9\times {10}^9\times 2.0\times {10}^{-6}\times \left(-3.5\times {10}^{-6}\right)}{0.25}\\ =-0.252J\end{array}$

Hence, the potential energy is.

Using conservation law:

$$\begin{gathered} K_a+U_a=K_b+U_b\begin{array}{c}\\ \end{array} \\ {K}_a=-U_a=0.252J \end{gathered}$$

Here $K_b$is zero and (v) is also zero because the potential is inversely proportional to (r) therefore $K_b$is

$$\begin{gathered} K_a=\frac{1}{2}m{v}^2 \\ =0.252 \end{gathered}$$

So,

$\begin{array}{l}{v}_a=\sqrt{\frac{2\times 0.25}{1.5\times {10}^{-3}}}\\ =18.3mls\end{array}$

Therefore, the initial speed is$18.3mls$.