Force exert on the other sphere.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 16, Problem 1OQ

To determine

Force exert on the other sphere.

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Force exert on the other sphere would become one quarter as large.:

Explanation of Solution

Given:

Charge on both the sphere is same, $q_{1} = q_{2} = q$

Initial separation distance, $r$

Final separation distance, $r_{2} = 2 r$

Formula used:

The force on the two charges is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$

Here,

$q_{1} q_{2}$ are the charges

$r$ is the separation distance

$ε_{o}$ is the electric permeability

Calculation:

Initial force on the sphere is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$ …… (1)

Final force on the sphere is calculated as

$\begin{matrix} F' = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{2 r^{2}} \\ = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}} \end{matrix}$ …… (2)

Divide Eq. (1)by Eq. (2)

$\begin{matrix} \frac{F}{F'} = \frac{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}}{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}}} \\ F' = \frac{F}{4} \end{matrix}$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

You want to fabricate a soft microfluidic chip like the one below. How would you go about fabricating this chip knowing that you are targeting a channel with a square cross-sectional profile of 200 μm by 200 μm. What materials and steps would you use and why? Disregard the process to form the inlet and outlet. Square Cross Section

1. What are the key steps involved in the fabrication of a semiconductor device. 2. You are hired by a chip manufacturing company, and you are asked to prepare a silicon wafer with the pattern below. Describe the process you would use. High Aspect Ratio Trenches Undoped Si Wafer P-doped Si 3. You would like to deposit material within a high aspect ratio trench. What approach would you use and why? 4. A person is setting up a small clean room space to carry out an outreach activity to educate high school students about patterning using photolithography. They obtained a positive photoresist, a used spin coater, a high energy light lamp for exposure and ordered a plastic transparency mask with a pattern on it to reduce cost. Upon trying this set up multiple times they find that the full resist gets developed, and they are unable to transfer the pattern onto the resist. Help them troubleshoot and find out why pattern of transfer has not been successful. 5. You are given a composite…

Two complex values are z1=8 + 8i, z2=15 + 7 i. z1∗ and z2∗ are the complex conjugate values. Any complex value can be expessed in the form of a+bi=reiθ. Find r and θ for (z1-z∗2)/z1+z2∗. Find r and θ for (z1−z2∗)z1z2∗ Please show all steps

Answer 2

Question

Chapter 16, Problem 1OQ

To determine

Force exert on the other sphere.

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Force exert on the other sphere would become one quarter as large.:

Explanation of Solution

Given:

Charge on both the sphere is same, $q_{1} = q_{2} = q$

Initial separation distance, $r$

Final separation distance, $r_{2} = 2 r$

Formula used:

The force on the two charges is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$

Here,

$q_{1} q_{2}$ are the charges

$r$ is the separation distance

$ε_{o}$ is the electric permeability

Calculation:

Initial force on the sphere is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$ …… (1)

Final force on the sphere is calculated as

$\begin{matrix} F' = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{2 r^{2}} \\ = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}} \end{matrix}$ …… (2)

Divide Eq. (1)by Eq. (2)

$\begin{matrix} \frac{F}{F'} = \frac{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}}{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}}} \\ F' = \frac{F}{4} \end{matrix}$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 16, Problem 1OQ

To determine

Force exert on the other sphere.

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Force exert on the other sphere would become one quarter as large.:

Explanation of Solution

Given:

Charge on both the sphere is same, $q_{1} = q_{2} = q$

Initial separation distance, $r$

Final separation distance, $r_{2} = 2 r$

Formula used:

The force on the two charges is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$

Here,

$q_{1} q_{2}$ are the charges

$r$ is the separation distance

$ε_{o}$ is the electric permeability

Calculation:

Initial force on the sphere is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$ …… (1)

Final force on the sphere is calculated as

$\begin{matrix} F' = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{2 r^{2}} \\ = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}} \end{matrix}$ …… (2)

Divide Eq. (1)by Eq. (2)

$\begin{matrix} \frac{F}{F'} = \frac{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}}{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}}} \\ F' = \frac{F}{4} \end{matrix}$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 16, Problem 1OQ

To determine

Force exert on the other sphere.

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Force exert on the other sphere would become one quarter as large.:

Explanation of Solution

Given:

Charge on both the sphere is same, $q_{1} = q_{2} = q$

Initial separation distance, $r$

Final separation distance, $r_{2} = 2 r$

Formula used:

The force on the two charges is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$

Here,

$q_{1} q_{2}$ are the charges

$r$ is the separation distance

$ε_{o}$ is the electric permeability

Calculation:

Initial force on the sphere is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$ …… (1)

Final force on the sphere is calculated as

$\begin{matrix} F' = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{2 r^{2}} \\ = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}} \end{matrix}$ …… (2)

Divide Eq. (1)by Eq. (2)

$\begin{matrix} \frac{F}{F'} = \frac{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}}{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}}} \\ F' = \frac{F}{4} \end{matrix}$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 16, Problem 1OQ

To determine

Force exert on the other sphere.

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Force exert on the other sphere would become one quarter as large.:

Explanation of Solution

Given:

Charge on both the sphere is same, $q_{1} = q_{2} = q$

Initial separation distance, $r$

Final separation distance, $r_{2} = 2 r$

Formula used:

The force on the two charges is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$

Here,

$q_{1} q_{2}$ are the charges

$r$ is the separation distance

$ε_{o}$ is the electric permeability

Calculation:

Initial force on the sphere is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$ …… (1)

Final force on the sphere is calculated as

$\begin{matrix} F' = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{2 r^{2}} \\ = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}} \end{matrix}$ …… (2)

Divide Eq. (1)by Eq. (2)

$\begin{matrix} \frac{F}{F'} = \frac{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}}{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}}} \\ F' = \frac{F}{4} \end{matrix}$

Answer 6

Chapter 16, Problem 1OQ

To determine

Force exert on the other sphere.

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Force exert on the other sphere would become one quarter as large.:

Explanation of Solution

Given:

Charge on both the sphere is same, $q_{1} = q_{2} = q$

Initial separation distance, $r$

Final separation distance, $r_{2} = 2 r$

Formula used:

The force on the two charges is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$

Here,

$q_{1} q_{2}$ are the charges

$r$ is the separation distance

$ε_{o}$ is the electric permeability

Calculation:

Initial force on the sphere is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$ …… (1)

Final force on the sphere is calculated as

$\begin{matrix} F' = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{2 r^{2}} \\ = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}} \end{matrix}$ …… (2)

Divide Eq. (1)by Eq. (2)

$\begin{matrix} \frac{F}{F'} = \frac{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}}{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}}} \\ F' = \frac{F}{4} \end{matrix}$

Answer 7

Chapter 16, Problem 1OQ

To determine

Force exert on the other sphere.

Answer 8

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Force exert on the other sphere would become one quarter as large.:

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

Charge on both the sphere is same, $q_{1} = q_{2} = q$

Initial separation distance, $r$

Final separation distance, $r_{2} = 2 r$

Formula used:

The force on the two charges is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$

Here,

$q_{1} q_{2}$ are the charges

$r$ is the separation distance

$ε_{o}$ is the electric permeability

Calculation:

Initial force on the sphere is given as

$F = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}$ …… (1)

Final force on the sphere is calculated as

$\begin{matrix} F' = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{2 r^{2}} \\ = \frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}} \end{matrix}$ …… (2)

Divide Eq. (1)by Eq. (2)

$\begin{matrix} \frac{F}{F'} = \frac{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{r^{2}}}{\frac{1}{4 π ε_{o}} \times \frac{q_{1} q_{2}}{4 r^{2}}} \\ F' = \frac{F}{4} \end{matrix}$