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Found in: Page 713

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

Two tiny metal spheres A and B, mass ${{\mathbf{m}}}_{{\mathbf{A}}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{g}}$ and ${{\mathbf{m}}}_{\mathbf{B}}{\mathbf{=}}{\mathbf{10}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{g}}$, have equal positive charge ${\mathbf{q}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{\mu C}}$. The spheres are connected by a mass less non-conducting string of length d=1.00 m, which is much greater than the radii of the spheres. (a) What is the electric potential energy of the system? (b) Suppose you cut the string. At that instant, what is the acceleration of each sphere? (c) A long time after you cut the string, what is the speed of each sphere?

1. The electric potential energy of the system is 0.225 J.
2. The acceleration of each sphere is 45 m/s2.
3. The speed of each sphere are 7.75 m/s and 3.87 m/s.
See the step by step solution

Step 1: The given data

1. Mass of the sphere A, ${\mathrm{m}}_{\mathrm{A}}=5.00\mathrm{g}\mathrm{or}5×{10}^{-3}\mathrm{kg}$
2. Mass of the sphere data-custom-editor="chemistry" ${\mathrm{m}}_{\mathrm{B}}=10.00\mathrm{g}\mathrm{or}10×{10}^{-3}\mathrm{kg}$
3. Charges: data-custom-editor="chemistry" ${\mathrm{q}}_{\mathrm{A}}={\mathrm{q}}_{\mathrm{B}}=5.0\mathrm{\mu C}$
4. Distance between the centers of the spheres, data-custom-editor="chemistry" $\mathrm{d}=1.00\mathrm{m}$.

Step 2: Understanding the concept of energy

Using the concept of energy, we can get the required value of the electric potential energy of the system by multiplying the force value with the separation between the particles. So, the force of Coulomb's law can be calculated using the given data. Again, using this force value, we can get the value of the acceleration of the system. Using the conservation law of momentum, we can get the velocity expression of sphere B. Now, using the concept of potential energy equal to the kinetic energy of the system, we get the value of the speed of sphere A.

Formulae:

The force due to Coulomb’s law, $\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{{\mathrm{q}}_{\mathrm{A}}{\mathrm{q}}_{\mathrm{B}}}{{\mathrm{d}}^{2}}$ (i)

The force due to Newton’s second law, data-custom-editor="chemistry" $\mathrm{F}=\mathrm{ma}$ (ii)

The momemtum value of a body in motion, data-custom-editor="chemistry" $\mathrm{p}=\mathrm{mv}$ (iii)

Step 3: a) Calculation of the electric potential energy of the system

Now, substituting all the known values in the equation (i), we can get the force value of the system as follows:

$\begin{array}{rcl}\mathrm{F}& =& \frac{1}{\left(4×3.14×8.85×{10}^{-12}\right)}\frac{\left(5×{10}^{-6}×5×{10}^{-6}\right)}{{1}^{2}}\\ & =& \left(9×{10}^{9}\right)\frac{\left(25×{10}^{-12}\right)}{{1}^{2}}\\ & =& 255×{10}^{-3}\\ & =& 0.255\mathrm{N}\end{array}$

This is the repulsive force acting between the metal spheres A and B. For r=1 m, the energy value is same as the force value.

Hence, the value of the energy is 0.255 J.

Step 4: b) Calculation of the acceleration of each sphere

Using the given values in equation (ii), we can get the acceleration of each sphere as follows: (from the above force value)

$\begin{array}{rcl}{\mathrm{a}}_{\mathrm{A}}& =& \frac{0.225}{{\mathrm{m}}_{\mathrm{A}}}\\ & =& \frac{0.225}{5×{10}^{-3}}\\ & =& 45\mathrm{m}/{\mathrm{s}}^{2}\end{array}$

Hence, the value of the acceleration of each sphere is 45 m/s2.

Step 5: c) Calculation of the speed of each sphere

Energy is conserved. The initial potential energy is 0.225 J, as calculated in part (a). The initial kinetic energy is zero since the spheres start from rest. The final potential energy is zero since the spheres are then far apart. The final kinetic energy is $\frac{1}{2}{\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}^{2}+\frac{1}{2}{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}^{2}$, where vA and vB are the final velocities.

Thus, the electric potential energy of the system as follows:

data-custom-editor="chemistry" $\mathrm{U}=\frac{1}{2}{\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}^{2}+\frac{1}{2}{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}^{2}..............\left(\mathrm{a}\right)$

Again, momentum is also conserved, so it is given as:

data-custom-editor="chemistry" $0={\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{\mathrm{B}}=-\left({\mathrm{m}}_{\mathrm{A}}/{\mathrm{m}}_{\mathrm{B}}\right){\mathrm{v}}_{\mathrm{B}}..............\left(\mathrm{b}\right)$

Thus, using the above value in equation (a) of the potential energy, we get the speed of sphere A as follows:

$\begin{array}{rcl}U& =& \frac{1}{2}\left({m}_{A}/{m}_{B}\right)\left({m}_{A}+{m}_{B}\right){v}_{A}^{2}\\ {\mathrm{v}}_{\mathrm{A}}& =& \sqrt{\frac{2{\mathrm{Um}}_{\mathrm{B}}}{{\mathrm{m}}_{\mathrm{A}}\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)}}\\ & =& \sqrt{\frac{2\left(0.255\mathrm{J}\right)\left(10×{10}^{-3}\mathrm{kg}\right)}{\left(5.0×{10}^{-3}\mathrm{kg}\right)\left(5.0×{10}^{-3}\mathrm{kg}+10.0×{10}^{-3}\mathrm{kg}\right)}}\\ & =& 7.75\mathrm{m}/\mathrm{s}\end{array}$

Now, using this value in equation (b), we get the value of the speed of the sphere B as follows:

$\begin{array}{rcl}{\mathrm{v}}_{\mathrm{B}}& =& -\left(\frac{5.0×{10}^{-3}\mathrm{kg}}{10.0×{10}^{-3}\mathrm{kg}}\right)\left(7.75\mathrm{m}/\mathrm{s}\right)\\ & =& -3.87\mathrm{m}/\mathrm{s}\\ \left|{\mathrm{v}}_{\mathrm{B}}\right|& =& 3.87\mathrm{m}/\mathrm{s}\end{array}$

Hence, the value of the speed of spheres A and B are 7.75 m/s and 3.87 m/s respectively.