Suggested languages for you:

Americas

Europe

37P

Expert-verified
Found in: Page 125

### Matter & Interactions

Book edition 4th edition
Author(s) Ruth W. Chabay, Bruce A. Sherwood
Pages 1135 pages
ISBN 9781118875865

# Two thin hollow plastic spheres, about the size of a ping-pong ball with masses (${{\mathbf{m}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{m}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{3}}{\mathbf{}}{\mathbf{kg}}$), have been rubbed with wool. Sphere 1 has a charge of ${{\mathbf{q}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{2}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{9}}{\mathbf{}}{\mathbf{C}}$ and is at location $\left(0.50,-0.20,0\right)$. Sphere 2 has a charge of ${{\mathbf{q}}}_{\mathbf{2}}{\mathbf{=}}{\mathbf{-}}{\mathbf{4}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{9}}{\mathbf{}}{\mathbf{C}}$ and is atlocation $\left(-0.40,0.40,0\right)$. It will be useful to draw a diagram of the situation, including the relevant vectors.a) What is the relative position vector $\stackrel{\mathbf{\to }}{\mathbf{r}}$ pointing from q1 to q2? b) What is the distance between q1 and q2? c) What is the unit vector $\stackrel{\mathbf{^}}{\mathbf{r}}$ in the direction of $\stackrel{\mathbf{\to }}{\mathbf{r}}$? 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 ${\mathbf{4}}\stackrel{\mathbf{\to }}{\mathbf{r}}$), 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 $\left(-0.90\mathrm{m}\right)\stackrel{^}{\mathrm{i}}+\left(+0.60\mathrm{m}\right)\stackrel{^}{\mathrm{j}}+0\stackrel{^}{\mathrm{k}}$.
• b) the distance between the points ${\mathrm{q}}_{1}\mathrm{and}{\mathrm{q}}_{2}$ is $1.081\mathrm{m}$.
• c) the unit vector is $\left(-0.832\mathrm{m}\right)\stackrel{^}{\mathrm{i}}+\left(0.555\mathrm{m}\right)\stackrel{^}{\mathrm{j}}$.
• d) the magnitude of the gravitational force is $2.283×{10}^{-16}\mathrm{N}$.
• e) the (vector) gravitational force exerted on ${\mathrm{q}}_{2}\mathrm{by}{\mathrm{q}}_{1}$is $\left(-1.899×{10}^{-16}\mathrm{N}\right)\stackrel{^}{\mathrm{i}}+\left(1.266×{10}^{-16}\mathrm{N}\right)\stackrel{^}{\mathrm{j}}$.
• f) the magnitude of the electric force exerted on ${\mathrm{q}}_{2}\mathrm{by}{\mathrm{q}}_{1}$ is $6.154×{10}^{-8}\mathrm{N}$.
• g) the (vector) electric force exerted on ${\mathrm{q}}_{2}\mathrm{by}{\mathrm{q}}_{1}$ is role="math" localid="1661249258040" $\left(5.12×{10}^{-9}\mathrm{N}\right)\stackrel{^}{\mathrm{i}}+\left(3.415×{10}^{-9}\mathrm{N}\right)\stackrel{^}{\mathrm{j}}$.
• h) the ratio of the magnitude of the electric force to the magnitude of the gravitational force is $2.69×{10}^{8}$.
• i) The ratio of the magnitude of the electric force and the gravitational force is $2.69×{10}^{8}$.
See the step by step solution

## Step 1: Identification of the given data

The given data can be listed below as:

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

## 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:

$\stackrel{\to }{{\mathrm{r}}_{1}}=\left(0.50\mathrm{m}\right)\stackrel{^}{\mathrm{i}}-\left(0.20\mathrm{m}\right)\stackrel{^}{\mathrm{j}}+\left(0\right)\stackrel{^}{\mathrm{k}}$

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

$\stackrel{\to }{{\mathrm{r}}_{2}}=-\left(0.40\mathrm{m}\right)\stackrel{^}{\mathrm{i}}+\left(0.40\mathrm{m}\right)\stackrel{^}{\mathrm{j}}+\left(0\mathrm{m}\right)\stackrel{^}{\mathrm{k}}$

Hence, the position vector from is described as:

$\stackrel{\to }{\mathrm{r}}=\stackrel{\to }{{\mathrm{r}}_{2}}-\stackrel{\to }{{\mathrm{r}}_{1}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{r}}=\left(-\left(0.40\mathrm{m}\right)\stackrel{^}{\mathrm{i}}+\left(0.40\mathrm{m}\right)\stackrel{^}{\mathrm{j}}+\left(0\mathrm{m}\right)\stackrel{^}{\mathrm{k}}\right)-\left(\left(0.50\mathrm{m}\right)\stackrel{^}{\mathrm{i}}-\left(0.20\mathrm{m}\right)\stackrel{^}{\mathrm{j}}+\left(0\mathrm{m}\right)\stackrel{^}{\mathrm{k}}\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{r}}=-\left(0.90\mathrm{m}\right)\stackrel{^}{\mathrm{i}}+\left(0.60\mathrm{m}\right)\stackrel{^}{\mathrm{j}}+\left(0\mathrm{m}\right)\stackrel{^}{\mathrm{k}}$

Thus, the relative position vector is $-\left(0.90\mathrm{m}\right)\stackrel{^}{\mathrm{i}}+\left(0.60\mathrm{m}\right)\stackrel{^}{\mathrm{j}}+\left(0\mathrm{m}\right)\stackrel{^}{\mathrm{k}}$.

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

Hence, the distance between the points ${\mathrm{q}}_{1}\mathrm{and}{\mathrm{q}}_{2}$ can be expressed as:

$\left|\stackrel{\to }{\mathrm{r}}\right|=\sqrt{{\left({\mathrm{r}}_{\mathrm{x}}\right)}^{2}+{\left({\mathrm{r}}_{\mathrm{y}}\right)}^{2}+{\left({\mathrm{r}}_{\mathrm{z}}\right)}^{2}}\phantom{\rule{0ex}{0ex}}$

Substituting the values in the above equation, we get-

$\left|\stackrel{\to }{\mathrm{r}}\right|=\sqrt{{\left(0.90\mathrm{m}\right)}^{2}+{\left(0.60\mathrm{m}\right)}^{2}+{\left(0\mathrm{m}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{\mathrm{r}}\right|=1.081\mathrm{m}$

Thus, the distance between the points ${\mathrm{q}}_{1}\mathrm{and}{\mathrm{q}}_{2}$ is $1.081\mathrm{m}$.

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

$\stackrel{^}{\mathrm{r}}=\frac{\stackrel{\to }{\mathrm{r}}}{\left|\stackrel{\to }{\mathrm{r}}\right|}$

Here, $\stackrel{^}{\mathrm{r}}$ is the unit vector, $\stackrel{\to }{\mathrm{r}}$ is the position vector, and $\left|\stackrel{\to }{\mathrm{r}}\right|$ is the distance between the points ${\mathrm{q}}_{1}\mathrm{and}{\mathrm{q}}_{2}$.

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

$\stackrel{^}{\mathrm{r}}=\frac{\left(-0.90\mathrm{m}\right)\stackrel{^}{\mathrm{i}}+\left(0.60\mathrm{m}\right)\stackrel{^}{\mathrm{j}}+\left(0\mathrm{m}\right)\stackrel{^}{\mathrm{k}}}{1.081\mathrm{m}}\phantom{\rule{0ex}{0ex}}\stackrel{^}{\mathrm{r}}=\left(-0.832\right)\stackrel{^}{\mathrm{i}}+\left(0.555\right)\stackrel{^}{\mathrm{j}}+\left(0\right)\stackrel{^}{\mathrm{k}}$

Thus, the unit vector is $\left(-0.832\right)\stackrel{^}{\mathrm{i}}+\left(0.555\right)\stackrel{^}{\mathrm{j}}$.

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

$\mathrm{F}=\frac{{\mathrm{Gm}}_{1}{\mathrm{m}}_{2}}{{\mathrm{r}}^{2}}............\left(2\right)$

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

Substituting the values in the above equation, we get-

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

Thus, the magnitude of the gravitational force is .

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

$\stackrel{\to }{\mathrm{F}}=\frac{{\mathrm{Gm}}_{1}{\mathrm{m}}_{2}}{{\mathrm{r}}^{2}}\stackrel{^}{\mathrm{r}}$

Here, $\stackrel{^}{\mathrm{r}}$ is the unit vector exerted on ${\mathrm{q}}_{2}\mathrm{by}{\mathrm{q}}_{1}$ .

Substituting the values in the above equation, we get-

$\stackrel{\to }{\mathrm{F}}=2.283×{10}^{-16}\mathrm{N}×\left(\left(-0.832\right)\stackrel{^}{\mathrm{i}}+\left(0.555\right)\stackrel{^}{\mathrm{j}}\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{F}}=\left(-1.899×{10}^{-16}\mathrm{N}\right)\stackrel{^}{\mathrm{i}}+\left(1.266×{10}^{-16}\mathrm{N}\right)\stackrel{^}{\mathrm{j}}$

Thus, the (vector) gravitational force exerted on ${\mathrm{q}}_{2}\mathrm{by}{\mathrm{q}}_{1}$ is $\left(-1.899×{10}^{-16}\mathrm{N}\right)\stackrel{^}{\mathrm{i}}+\left(1.266×{10}^{-16}\mathrm{N}\right)\stackrel{^}{\mathrm{j}}$.

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

$\mathrm{F}=\mathrm{k}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}^{2}}.......\left(3\right)$

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

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

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

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

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

$\stackrel{\to }{\mathrm{F}}=\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}^{2}}\stackrel{^}{\mathrm{r}}$

Here, $\stackrel{^}{\mathrm{r}}$ is the unit vector exerted on ${\mathrm{q}}_{2}\mathrm{by}{\mathrm{q}}_{1}$.

Substituting the values in the above equation, we get-

$\stackrel{\to }{\mathrm{F}}=6.154×{10}^{-9}\mathrm{N}×\left(-0.832\right)\stackrel{^}{\mathrm{i}}+\left(0.555\right)\stackrel{^}{\mathrm{j}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{F}}=\left(5.12×{10}^{-9}\mathrm{N}\right)\stackrel{^}{\mathrm{i}}+\left(3.145×{10}^{-9}\right)\stackrel{^}{\mathrm{j}}$

Thus, the (vector) electric force exerted on ${\mathrm{q}}_{2}\mathrm{by}{\mathrm{q}}_{1}$ is $\left(5.12×{10}^{-9}\mathrm{N}\right)\stackrel{^}{\mathrm{i}}+\left(3.415×{10}^{-9}\right)\stackrel{^}{\mathrm{j}}$.

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

$\mathrm{ratio}=\frac{6.154×{10}^{-8}\mathrm{N}}{2.283×{10}^{-16}\mathrm{N}}\phantom{\rule{0ex}{0ex}}\mathrm{ratio}=2.69×{10}^{8}\mathrm{N}$

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

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

$\left|\stackrel{\to }{\mathrm{r}}\right|=4×1.081\mathrm{m}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{\mathrm{r}}\right|=4.324\mathrm{m}$

Per equation (3) and for the distance 4$\stackrel{\to }{\mathrm{r}}$ and charges ${\mathrm{q}}_{2}\mathrm{and}{\mathrm{q}}_{1}$ , the magnitude of the electric force will be:

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

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

$\mathrm{F}=\frac{\left(6.67×{10}^{-11}\mathrm{N}.{\mathrm{m}}^{2}/{\mathrm{kg}}^{2}\right)×{\left(2×{10}^{-3}\mathrm{kg}\right)}^{2}}{{\left(4.324\mathrm{m}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{F}=\left(\frac{6.67×{10}^{-11}×{\left(2×{10}^{-3}\right)}^{2}}{18.696}\right)\left(\frac{1\mathrm{N}.{\mathrm{m}}^{2}/{\mathrm{kg}}^{2}×1{\mathrm{kg}}^{2}}{1{\mathrm{m}}^{2}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{F}=1.4269×{10}^{-17}.\left(1\mathrm{N}\right)\phantom{\rule{0ex}{0ex}}\mathrm{F}=1.4269×{10}^{-17}\mathrm{N}$

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

$\mathrm{ratio}=\frac{3.846×{10}^{-9}\mathrm{N}}{1.4269×{10}^{-17}\mathrm{N}}\phantom{\rule{0ex}{0ex}}\mathrm{ratio}=2.69×{10}^{8}$

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