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

A block of mass m = 7.00 kg is released from rest from point and slides on the frictionless track shown in the figure below. (Assume h₂ = 7.80 m.) a m ha 3.20 m 2.00 m i (a) Determine the block's speed at points ® and point B ©. m/s m/s point (b) Determine the net work done by the gravitational force on the block as it moves from point J A to point

A 1.10 x 10²-g particle is released from rest at point A on the inside of a smooth hemispherical bowl of radius R R B 2R/3 (a) Calculate its gravitational potential energy at A relative to B. ] (b) Calculate its kinetic energy at B. ] (c) Calculate its speed at B. m/s (d) Calculate its potential energy at C relative to B. J (e) Calculate its kinetic energy at C. ] = 26.5 cm (figure below).

Report on the percentage errors (with uncertainty) between the value of 'k' from the F vs displacement plot and each of the values of 'k' from the period measurements. Please comment on the goodness of the results. Value of k = Spring constant k = 50.00 N/m Each of the values of k from period measurements: Six Measurements of time for 5 osccilations: t1 = 7.76s, t2=8.00s, t3=7.40s, t4=7.00s, t5=6.90s, t6=7.10s (t1-tavg)^2 = (7.76-7.36)^2 = 0.16%(t2-tavg)^2 =(8.00-7.36)^2 = 0.4096%(t3-tavg)^2 =(7.40-7.36)^2 = 0.0016%(t4-tavg)^2 =(7.00-7.36)^2 = 0.1296%(t5-tavg)^2 =(6.90-7.36)^2 = 0.2116%(t6-tavg)^2 =(7.10-7.36)^2 = 0.0676

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)

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