Book edition 4th edition

Author(s) Ruth W. Chabay, Bruce A. Sherwood

Pages 1135 pages

ISBN 9781118875865

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Short Answer

Two thin hollow plastic spheres, about the size of a ping-pong ball with masses ( $m_{1} = m_{2} = 2 \times 10^{- 3} kg$ ), have been rubbed with wool. Sphere 1 has a charge of $q_{1} = - 2 \times 10^{- 9} C$ and is at location $(0.50, - 0.20, 0)$ . Sphere 2 has a charge of $q_{2} = - 4 \times 10^{- 9} C$ and is atlocation $(- 0.40, 0.40, 0)$ . It will be useful to draw a diagram of the situation, including the relevant vectors.
a) What is the relative position vector $\vec{r}$ pointing from q₁ to q₂? b) What is the distance between q₁ and q₂? c) What is the unit vector $\hat{r}$ in the direction of $\vec{r}$ ? d) What is the magnitude of the gravitational force exerted on q₂ by q₁? e) What is the (vector) gravitational force exerted on q₂ by q₁? f) What is the magnitude of the electric force exerted on q₂ by q₁? g) What is the (vector) electric force exerted on q₂ by q₁? h) What is the ratio of the magnitude of the electric force to the magnitude of the gravitational force? i) if the two masses were four times further away (that is, if the distance between the masses were $4 \vec{r}$ ), what would be the ratio of the magnitude of the electric force to the magnitude of the gravitational force now?

a) The relative position vector is $(- 0.90 m) \hat{i} + (+ 0.60 m) \hat{j} + 0 \hat{k}$ .
b) the distance between the points $q_{1} and q_{2}$ is $1.081 m$ .
c) the unit vector is $(- 0.832 m) \hat{i} + (0.555 m) \hat{j}$ .
d) the magnitude of the gravitational force is $2.283 \times 10^{- 16} N$ .
e) the (vector) gravitational force exerted on $q_{2} by q_{1}$ is $(- 1.899 \times 10^{- 16} N) \hat{i} + (1.266 \times 10^{- 16} N) \hat{j}$ .
f) the magnitude of the electric force exerted on $q_{2} by q_{1}$ is $6.154 \times 10^{- 8} N$ .
g) the (vector) electric force exerted on $q_{2} by q_{1}$ is role="math" localid="1661249258040" $(5.12 \times 10^{- 9} N) \hat{i} + (3.415 \times 10^{- 9} N) \hat{j}$ .
h) the ratio of the magnitude of the electric force to the magnitude of the gravitational force is $2.69 \times 10^{8}$ .
i) The ratio of the magnitude of the electric force and the gravitational force is $2.69 \times 10^{8}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Identification of the given data

The given data can be listed below as:

The mass of the hollow plastic spheres is $2 \times 10^{- 3} kg$ .
The charge of sphere 1 is $- 2 \times 10^{- 9} C$ .
The first sphere is at the location $(0.50, - 0.20, 0)$ .
The charge of sphere 2 is $- 4 \times 10^{- 9} C$ .
The first sphere is at the location $(- 0.40, 0.40, 0)$ .

Step 2: Significance of the Newton’s gravitational NS Coulomb’s law in identifying the forces

Newton’s gravitational law states that the particle in the field of another particle feels a force that is directly proportional to the product of the mass and inversely proportional to the square of the distance.

Coulomb’s law states that the unlike charges attract and like charges repel each other.

The equation of the gravitational and Coulomb’s equation give the gravitational and the electric force along with the position, distance, unit vector, and the ratio of the gravitational and electric force.

Step 3: Determination of the gravitational and the electric force along with the position and the unit vector and the ratio of the electric and gravitational forces

The free-body diagram of the two particles and the vectors have been provided below.

a) According to Newton’s laws, the position vector for the sphere 1 can be expressed as:

$\vec{r_{1}} = (0.50 m) \hat{i} - (0.20 m) \hat{j} + (0) \hat{k}$

Similarly, the position vector for the sphere 2 can be expressed as:

$\vec{r_{2}} = - (0.40 m) \hat{i} + (0.40 m) \hat{j} + (0 m) \hat{k}$

Hence, the position vector from is described as:

$\vec{r} = \vec{r_{2}} - \vec{r_{1}} \vec{r} = (- (0.40 m) \hat{i} + (0.40 m) \hat{j} + (0 m) \hat{k}) - ((0.50 m) \hat{i} - (0.20 m) \hat{j} + (0 m) \hat{k}) \vec{r} = - (0.90 m) \hat{i} + (0.60 m) \hat{j} + (0 m) \hat{k}$

Thus, the relative position vector is $- (0.90 m) \hat{i} + (0.60 m) \hat{j} + (0 m) \hat{k}$ .

b) The position vector can help calculate the distance between the particles.

Hence, the distance between the points $q_{1} and q_{2}$ can be expressed as:

$|\vec{r}| = \sqrt{{(r_{x})}^{2} + {(r_{y})}^{2} + {(r_{z})}^{2}}$

Substituting the values in the above equation, we get-

$|\vec{r}| = \sqrt{{(0.90 m)}^{2} + {(0.60 m)}^{2} + {(0 m)}^{2}} |\vec{r}| = 1.081 m$

Thus, the distance between the points $q_{1} and q_{2}$ is $1.081 m$ .

c) From Newton’s gravitational law, the unit vector can be expressed as:

$\hat{r} = \frac{\vec{r}}{|\vec{r}|}$

Here, $\hat{r}$ is the unit vector, $\vec{r}$ is the position vector, and $|\vec{r}|$ is the distance between the points $q_{1} and q_{2}$ .

Substituting the values in the above equation and using the value from the equation , we get-

$\hat{r} = \frac{(- 0.90 m) \hat{i} + (0.60 m) \hat{j} + (0 m) \hat{k}}{1.081 m} \hat{r} = (- 0.832) \hat{i} + (0.555) \hat{j} + (0) \hat{k}$

Thus, the unit vector is $(- 0.832) \hat{i} + (0.555) \hat{j}$ .

d) From Newton’s gravitational law, the gravitational force exerted on the sphere q₂ can be expressed as:

$F = \frac{{Gm}_{1} m_{2}}{r^{2}} . . . . . . . . . . . . (2)$

Here, F is the gravitational force, G is the gravitational constant, $m_{1} and m_{2}$ are the mass of the spheres that are $2 \times 10^{- 3} kg$ , and r is the distance amongst them.

Substituting the values in the above equation, we get-

$\begin{array}{rcl} F & = & \frac{(6.67 \times 10^{- 11} N . m^{2} / {kg}^{2}) \times {(2 \times 10^{- 3} kg)}^{2}}{{(1.081 m)}^{2}} \\ = & (\frac{6.67 \times 10^{- 11} \times {(2 \times 10^{- 3})}^{2}}{{(1.081)}^{2}}) . (\frac{1 N . m^{2} / {kg}^{2} \times 1 kg}{1 m^{2}}) \\ = & \frac{26.68 \times 10^{- 17}}{1.168561} . (1 N) \\ F & = & 2.283 \times 10^{- 16} N \end{array}$

Thus, the magnitude of the gravitational force is .

e) From Newton’s gravitational law, the vector gravitational force exerted on $q_{2} and q_{1}$ is expressed as:

$\vec{F} = \frac{{Gm}_{1} m_{2}}{r^{2}} \hat{r}$

Here, $\hat{r}$ is the unit vector exerted on $q_{2} by q_{1}$ .

Substituting the values in the above equation, we get-

$\vec{F} = 2.283 \times 10^{- 16} N \times ((- 0.832) \hat{i} + (0.555) \hat{j}) \vec{F} = (- 1.899 \times 10^{- 16} N) \hat{i} + (1.266 \times 10^{- 16} N) \hat{j}$

Thus, the (vector) gravitational force exerted on $q_{2} by q_{1}$ is $(- 1.899 \times 10^{- 16} N) \hat{i} + (1.266 \times 10^{- 16} N) \hat{j}$ .

f) According to Coulomb’s law, the magnitude of the electric force can be expressed as:

$F = k \frac{q_{1} q_{2}}{r^{2}} . . . . . . . (3)$

Here, F is the magnitude of the electric force, k is the Coulomb’s constant that is about $8.99 \times 10^{9} N . m^{2} / C^{2}$ , $q_{1} and q_{2}$ are the charges of the masses, and r is the distance amongst them.

Substituting the values in the above equation (3), we get-

$\begin{array}{rcl} F & = & 8.99 \times 10^{9} N . m^{2} / C^{2} \times \frac{(- 2 \times 10^{- 9} C) \times (- 4 \times 10^{- 9} C)}{{(1.081 m)}^{2}} \\ = & (8.99 \times 10^{9} \times \frac{8 \times 10^{- 18}}{1.16586}) . (1 N . m^{2} / C^{2} \times \frac{1 C^{2}}{1 m^{2}}) \\ = & \frac{7.192 \times 10^{- 8}}{1.168561} . (1 N) \\ F & = & 6.154 \times 10^{- 8} N \end{array}$

Thus, the magnitude of the electric force exerted on $q_{2} by q_{1}$ is $6.154 \times 10^{- 8} N$ .

g) From Coulomb’s law, the vector electric force exerted on $q_{2} by q_{1}$ is expressed as:

$\vec{F} = \frac{q_{1} q_{2}}{r^{2}} \hat{r}$

Here, $\hat{r}$ is the unit vector exerted on $q_{2} by q_{1}$ .

Substituting the values in the above equation, we get-

$\vec{F} = 6.154 \times 10^{- 9} N \times (- 0.832) \hat{i} + (0.555) \hat{j} \vec{F} = (5.12 \times 10^{- 9} N) \hat{i} + (3.145 \times 10^{- 9}) \hat{j}$

Thus, the (vector) electric force exerted on $q_{2} by q_{1}$ is $(5.12 \times 10^{- 9} N) \hat{i} + (3.415 \times 10^{- 9}) \hat{j}$ .

h) The ratio of the magnitude of the electric force to the magnitude of the gravitational force can be expressed as:

$ratio = \frac{6.154 \times 10^{- 8} N}{2.283 \times 10^{- 16} N} ratio = 2.69 \times 10^{8} N$

Thus, the ratio of the magnitude of the electric force to the magnitude of the gravitational force is $2.69 \times 10^{8}$ .

i) If the distance between the masses becomes $4 \vec{r}$ , then the distance between the masses can be expressed from equation :

$|\vec{r}| = 4 \times 1.081 m |\vec{r}| = 4.324 m$

Per equation (3) and for the distance 4 $\vec{r}$ and charges $q_{2} and q_{1}$ , the magnitude of the electric force will be:

$\begin{array}{rcl} F & = & 8.99 \times 10^{9} N . m^{2} / C^{2} \times \frac{(- 2 \times 10^{- 9} C) \times (- 4 \times 10^{- 9} C)}{{(4.324 m)}^{2}} \\ = & (\frac{8.99 \times 10^{9} \times - 2 \times 10^{- 9} \times 4 \times 10^{- 9}}{18.696}) (\frac{1 N . m^{2} / C^{2} \times 1 C \times 1 C}{1 m^{2}}) \\ = & 8.99 \times 10^{9} \times \frac{8 \times 10^{- 18}}{18.696} . (1 N) \\ F & = & 3.846 \times 10^{- 9} N \end{array}$

From equation (2), for the distance $4 \vec{r}$ , the magnitude of the gravitational force can be calculated as:

$F = \frac{(6.67 \times 10^{- 11} N . m^{2} / {kg}^{2}) \times {(2 \times 10^{- 3} kg)}^{2}}{{(4.324 m)}^{2}} F = (\frac{6.67 \times 10^{- 11} \times {(2 \times 10^{- 3})}^{2}}{18.696}) (\frac{1 N . m^{2} / {kg}^{2} \times 1 {kg}^{2}}{1 m^{2}}) F = 1.4269 \times 10^{- 17} . (1 N) F = 1.4269 \times 10^{- 17} N$

Then, the ratio of the magnitude of the electric force and the gravitational force is-

$ratio = \frac{3.846 \times 10^{- 9} N}{1.4269 \times 10^{- 17} N} ratio = 2.69 \times 10^{8}$

Thus, the ratio of the magnitude of the electric force and the gravitational force is $2.69 \times 10^{8}$ .

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Short Answer

Step by Step Solution

Step 1: Identification of the given data

Step 2: Significance of the Newton’s gravitational NS Coulomb’s law in identifying the forces

Step 3: Determination of the gravitational and the electric force along with the position and the unit vector and the ratio of the electric and gravitational forces

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Short Answer

Step by Step Solution

Step 1: Identification of the given data

Step 2: Significance of the Newton’s gravitational NS Coulomb’s law in identifying the forces

Step 3: Determination of the gravitational and the electric force along with the position and the unit vector and the ratio of the electric and gravitational forces

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