Ask question

Ask question

All categories

2 years ago

8

Two charges are separted by a distance d and exert mutual attractive forces of f on each other. If the charges are separated by

a distate of 2d what are the new mutual forces?

1 answer:

ASHA 777 [7]2 years ago

3 0

The new force between the charges when the distance become 2d will be F'=4F

<h3>What is electrostatic force?</h3>

When two charged particles are separated by the distance d then the force of attraction or repultion acts on the charged particle depending upon the nature of the charge whether it is positive or negative.

The formula for the electrostatic force is given by

$F=\dfrac{kq_1q_2}{d^2}$

Now if the value of d becomes 2d then the formula will become:

$F'=\dfrac{kq_1q_2}{(2d)^2}=\dfrac{kq_1q_2}{4d}$

$\dfrac{F'}{F}=\dfrac{1}{4}$

$F=4F'$

Hence the new force between the charges when the distance become 2d will be F'=4F

To know more about electrostatic force follow

brainly.com/question/17692887

#SPJ4

You might be interested in

Which of the following is not a component of a normal experiment?

MaRussiya [10]

Answer:

D. Validity group

Explanation:

A validity group is not necessarily a component of a normal experiment.

A normal experiment is made up of the dependent variables, independent variables, and most times the control group.

The dependent variable is one that relies on the outcome of the effect produced from the experimental design
The independent variable is the one that causes a desired effect under the study.
The control group is a group that does not receive any treatment nor attenuation of the conditions.

4 0

3 years ago

G = 10 N/kg or 10 m/s2

Irina18 [472]

Answer:

a) $U_{g} = 40\,J$ , b) $\eta = 70\,\%$ , c) $v = 20\,\frac{m}{s}$

Explanation:

a) The initial potential energy is:

$U_{g} = (0.2\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (20\,m)$

$U_{g} = 40\,J$

b) The efficiency of the bounce is:

$\eta = \left(\frac{14\,m}{20\,m} \right)\times 100\,\%$

$\eta = 70\,\%$

c) The final speed of Danielle right before reaching the bottom of the hill is determined from the Principle of Energy Conservation:

$K = U_{g}$

$U_{g} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot U_{g}}{m} }$

$v = \sqrt{\frac{2\cdot (40\,J)}{0.2\,kg} }$

$v = 20\,\frac{m}{s}$

5 0

3 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

a)

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

b)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

c)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

4 years ago

Is there any change in the pressure of container filled with water when the volumed is increased

marshall27 [118]

Not really the volume of a container is simply length X width X depth so just how big the container unless the water is pressurized by some sort of weight or if the containers air pressure is lowered

7 0

4 years ago

The greatest ocean depths on the earth are found in the marianas trench near the philippines. calculate the pressure due to the

tensa zangetsu [6.8K]

First, let us derive our working equation. We all know that pressure is the force exerted on an area of space. In equation, that would be: P = F/A. From Newton's Law of Second Motion, force is equal to the product of mass and gravity: F = mg. So, we can substitute F to the first equation so that it becomes, P = mg/A. Now, pressure can also be determined as the force exerted by a fluid on an area. This fluid can be measure in terms of volume. Relating volume and mass, we use the parameter of density: ρ = m/V. Simplifying further in terms of height, Volume is the product of the cross-sectional area and the height. So, V = A*h. The working equation will then be derived to be:

P = ρgh

This type of pressure is called the hydrostatic pressure, the pressure exerted by the fluid over a known height. Next, we find the literature data of the density of seawater. From studies, seawater has a density ranging from 1,020 to 1,030 kg/m³. Let's just use 1,020 kg/m³. Substituting the values and making sure that the units are consistent:

P = (1,020 kg/m³)(9.81 m/s²)(11 km)*(1,000 m/1km)
P = 110,068,200 Pa or 110.07 MPa

6 0

3 years ago