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

explain how to find the answer

help and explain how to find the answer

What is the equivalent resistance between points A and B of the network shown in the figure? A • 12 B 4.0 6.0 Ω 8.0 Ω

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)

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