Chapter 25, Problem 1CQ

a. Charge q₁is distance r from a positive point charge Q. Charge q₂= q₁/3 is distance 2r from Q. What is the ratio U₁/U₂ of their potential energies due to their interactions with Q? b. Charge q₁is distance s from the negative plate of a parallel-plate capacitor. Charge q₂= q₁/3 is distance 2s from the negative plate. What is the ratio U₁/U₂ of their potential energies?

To determine

a. Ratio of potential energy $\left({\text{U}}_{\text{1}}{\text{/U}}_{\text{2}}\right)$

b. Ratio of potential energy $\left({\text{U}}_{\text{1}}{\text{/U}}_{\text{2}}\right)$

Answer to Problem 1CQ

Solution:

a. Ratio of the potential energy is $6/1$

b. Ratio of the potential energy is $6/1$

Explanation of Solution

Given:

(a)

Charge $q_1$ is at distance $r$ from the positive point charge $Q$

Charge $q_2={\text{q}}_{\text{1}}\text{/3}$ is at distance $2r$ from the positive point charge $Q$

(b)

Charge $q_1$ is at distance $s$ from the negative plate charge $-Q$

Charge $q_2={\text{q}}_{\text{1}}\text{/3}$ is at distance $2s$ from the negative plate charge $-Q$

Formula used:

The potential energy of the charge particle is given by the formula

$U=\frac{1}{4\pi {\epsilon }_o}\times \frac{q_{1}Q}{r}$

Here,

$U$ is the potential energy

${\epsilon }_o$ is electric permissibility

$q_1\&Q$ are charges

$r$ is the distance between charge particles

Calculation:

(a)

Consider the diagram of the charge particles

Figure.1

The potential energy of the charge particle $q_1$ is given as

$U_1=\frac{1}{4\pi {\epsilon }_o}\times \frac{q_{1}Q}{r}$ …… (1)

The potential energy of the charge particle $q_2$ is given as

$U_2=\frac{1}{4\pi {\epsilon }_o}\times \frac{q_{2}Q}{r}$

Plugging the values in the above equation

$U_2=\frac{1}{4\pi {\epsilon }_o}\times \frac{q_{1}Q}{3\left(2r\right)}$ …… (2)

Calculate the ratio of the potential energy by dividing Eq. (1) by Eq. (2)

$\begin{array}{c}\frac{U_1}{U_2}=\frac{\frac{1}{4\pi {\epsilon }_o}\times \frac{q_{1}Q}{r}}{\frac{1}{4\pi {\epsilon }_o}\times \frac{q_{1}Q}{3\left(2r\right)}}\\ \frac{U_1}{U_2}=\frac{6}{1}\end{array}$

Hence the ration of their potential energy is $6\text{:}1$.

(b)

Consider the diagram of the charge particles

Figure.1

The potential energy of the charge particle $q_1$ is given as

$U_1=\frac{1}{4\pi {\epsilon }_o}\times \frac{q_{1}Q}{r}$

Plugging the values in the above equation

$U_1=\frac{1}{4\pi {\epsilon }_o}\times \frac{q_1\left(-Q\right)}{s}$ …… (3)

The potential energy of the charge particle $q_2$ is given as

$U_2=\frac{1}{4\pi {\epsilon }_o}\times \frac{q_{2}Q}{r}$

Plugging the values in the above equation

$U_2=\frac{1}{4\pi {\epsilon }_o}\times \frac{q_1\left(-Q\right)}{3\left(2s\right)}$ …… (4)

Calculate the ratio of the potential energy by dividing Eq. (3) by Eq. (4)

$\begin{array}{c}\frac{U_1}{U_2}=\frac{\frac{1}{4\pi {\epsilon }_o}\times \frac{q_1\left(-Q\right)}{s}}{\frac{1}{4\pi {\epsilon }_o}\times \frac{q_1\left(-Q\right)}{3\left(2s\right)}}\\ \frac{U_1}{U_2}=\frac{6}{1}\end{array}$

Hence the ration of their potential energy is $6\text{:}1$.

Conclusion:

a. Ratio of the potential energy is $6/1$

b. Ratio of the potential energy is $6/1$