Force exert on the other sphere.

Question

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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)

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

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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)

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

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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)

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

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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)

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

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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)

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