All categories

3 years ago

Rearrange the Coulomb's Law formula to find distance(r)

2 answers:

6 0

Coulomb's law describes the magnitude of the electrostatic force between two electric charges. The Coulomb's law formula is:
 F = Ke * q1 * q2 / r2

Where:
 q1: Charge of object 1
 q2: Charge of object 2
 r: Distance between the two objects
 F: Force between the two objects. A positive force implies a repulsive interaction, while a negative force implies an attractive interaction
 Ke = Coulomb Constant, 8.9875517873681764 * 109 N.m2.C-2

Coulomb's Law Examples:
What is the magnitude of the force between two protons which are 1.6E10-6 meters apart?

The charge of 1 proton is +1e (+1.602E-19 C).

F = 8.9875517873681764 * 109 * 1.602E-19 * 1.602E-19 / (1.6E-6 * 1.6 E-6) = 9 * 10-17 N

Source- http://www.endmemo.com/physics/coulomb.php

f = k (q^1)(q^2)/r^2

both sides by r^2 Multiply
fr^2/r^2 = kq^3/r^2

both sides by r^2 divide
f = kq^3/r^2

Answer:
f = kq^3/r^2

Hope this helps!!!!

vaieri [72.5K]3 years ago

4 0

Answer:

distance in terms of force is given as

$r = \sqrt{\frac{kq_1q_2}{f}}$

Explanation:

here as we know that the force between two charges placed at distance "r" is given by Coulomb's law

It is given by the formula

$F = \frac{kq_1q_2}{r^2}$

now we have to rearrange it for finding the distance

so here we will multiply both sides by square of the distance

$f \times r^2 = kq_1q_2$

now we will divide both sides force "f"

$r^2 = \frac{kq_1q_2}{f}$

now square root both sides

$r = \sqrt{\frac{kq_1q_2}{f}}$

You might be interested in

An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place

anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

$mu=(M+m)v$

$v=\frac{m}{M+m}\times u$

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

$\frac{(M+m)v^2}{2}=(M+m)gh$

here $h=L(1-\cos \theta )$ from diagram

therefore

$v=\sqrt{2gL(1-\cos \theta )}$

initial velocity in terms of v

$u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}$

For first case $\theta =6.8^{\circ}$

$u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}$

for second case $\theta =11.4^{\circ}$

$u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}$

Therefore $\frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}$

$\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}$

$\frac{u_1}{u_2}=1.084$

i.e. $\frac{v_1}{v_2}=1.084$

4 0

2 years ago

A free-body diagram is a special example of the vector diagrams that will

Genrish500 [490]

From someone who has taken physics this is, True

6 0

3 years ago

In which of these samples do the molecules mist likely have the least kinetic energy

KiRa [710]

Provide the examples again, no samples are presents.

8 0

2 years ago

Water is falling on a surface, wetting a circular area that is expanding at a rate of 4 mm2 /s. How fast is the radius of the we

STALIN [3.7K]

Answer:

The radius of the wetter area expands at a rate of $4.244\times 10^{-3}$ milimeters per second when radius is 150 milimeters.

Explanation:

From Geometry we remember that area of a circle is described by this expression:

$A =\pi\cdot r^{2}$ (Eq. 1)

Where:

$r$ - Radius of the circle, measured in milimeters.

$A$ - Area of the circle, measured in square milimeters.

Then, the rate of change of the area in time is derived by concept of rate of change, that is:

$\frac{dA}{dt} = 2\pi\cdot r\cdot \frac{dr}{dt}$ (Eq. 2)

Where:

$\frac{dr}{dt}$ - Rate of change of radius in time, measured in milimeters per second.

$\frac{dA}{dt}$ - Rate of change of area in time, measured in square milimeters per second.

Now the rate of change of radius in time is cleared within equation above:

$\frac{dr}{dt} = \left(\frac{1}{2\pi\cdot r}\right)\cdot \frac{dA}{dt}$

If we know that $r = 150\,mm$ and $\frac{dA}{dt} = 4\,\frac{mm^{2}}{s}$ , then the rate of change of radius in time is:

$\frac{dr}{dt} = \left[\frac{1}{2\pi\cdot (150\,m)} \right] \cdot \left(4\,\frac{mm^{2}}{s} \right)$

$\frac{dr}{dt}\approx 4.244\times 10^{-3}\,\frac{mm}{s}$

The radius of the wetter area expands at a rate of $4.244\times 10^{-3}$ milimeters per second when radius is 150 milimeters.

7 0

3 years ago

What causes ray of white light to spilt into seven colors when passed through a prism

saul85 [17]

Dispersion, I think.

3 0

2 years ago

Read 2 more answers