Chapter 17, Problem 38QAP

To determine

(a)

The speed each cell would need when very far away from each other to get close enough to just touch.

Answer to Problem 38QAP

The speed each cell would need when very far away from each other to get close enough to just touch is $321.5\text{m/s}$.

Explanation of Solution

Given info:

Mass of each cell is, $m=9.0\times {10}^{-14}\text{kg}$.

Charge on first cell is, $q_1=-2.50\text{pC}$.

Charge on second cell is, $q_2=-3.10\text{pC}$.

Diameter of each cell is, $d=7.5\mu \text{m}$.

Formula used:

Formula for the electric potential energy for two point charges is,

  $U_{\text{elec}}=k\frac{q_1{q}_2}{d}$

Calculation:

When the two cells are close enough to just touch, their kinetic energy will be zero. So, they have only electric potential energy at this point.

When the two cells are very far away from each other, their electric potential energy will be zero. So, they have only kinetic energy in this situation.

The kinetic energy of each cell when they are very far away can be calculated as,

  $K=\frac{1}{2}m{v}^2$

The electric potential energy of two cells when they get close enough to just touch can be calculated as,

  $U_{\text{elec}}=k\frac{q_1{q}_2}{d}$

From the conservation of energy, we get

  $\begin{array}{c}2K=U_{elec}\\ 2\times \frac{1}{2}m{v}^2=k\frac{q_1{q}_2}{d}\\ v^2=k\frac{q_1{q}_2}{md}\end{array}$

Substituting the given values in the above equation, we get

  $\begin{array}{c}{v}^2=9\times {10}^9\times \frac{\left(-2.50\times {10}^{-12}\right)\left(-3.10\times {10}^{-12}\right)}{\left(9.0\times {10}^{-14}\right)\left(7.5\times {10}^{-6}\right)}\\ v^2=103333.33\\ v\approx 321.5\text{m/s}\end{array}$

Conclusion:

Thus, the speed each cell would need when very far away from each other to get close enough to just touch is $321.5\text{m/s}$.

To determine

(b)

The magnitude of the maximum acceleration of the each cell.

Answer to Problem 38QAP

The magnitude of the maximum acceleration of the each cell is $1.38\times {10}^{10}{\text{m/s}}^{\text{2}}$.

Explanation of Solution

Given info:

Mass of each cell is, $m=9.0\times {10}^{-14}\text{kg}$.

Charge on first cell is, $q_1=-2.50\text{pC}$.

Charge on second cell is, $q_2=-3.10\text{pC}$.

Diameter of each cell is, $d=7.5\mu \text{m}$.

Formula used:

The formula for the electrostatic force between two charged particles is given as,

  $F=k\frac{q_1{q}_2}{d^2}$

Calculation:

Substituting the given values in the above equation, we get

  $\begin{array}{c}F=9\times {10}^9\times \frac{\left(-2.50\times {10}^{-12}\right)\left(-3.10\times {10}^{-12}\right)}{{\left(7.5\times {10}^{-6}\right)}^2}\\ F=1.24\times {10}^{-3}\text{N}\end{array}$

The maximum acceleration of each cell can be calculated as,

  $\begin{array}{c}a=\frac{F}{m}\\ a=\frac{1.24\times {10}^{-3}}{9.0\times {10}^{-14}}\\ a=1.38\times {10}^{10}{\text{m/s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the magnitude of the maximum acceleration of the each cell is $1.38\times {10}^{10}{\text{m/s}}^{\text{2}}$.