Ask question

Ask question

All categories

3 years ago

12

If the charges attracting each other in the preceding problem have equal magnitudes,show that the magnitude of each charge is 2.

8

1 answer:

Schach [20]3 years ago

6 0

Answer:

The magnitude of each charge is $2.82\times10^{-6}\ C$

Explanation:

Suppose the two point charges are separated by 6 cm. The attractive force between them is 20 N.

We need to calculate the magnitude of each charge

Using formula of force

$F=\dfrac{kq^2}{r^2}$

Where, q = charge

r = separation

Put the value into the formula

$20=\dfrac{9\times10^{9}\times q^2}{(6\times10^{-2})^2}$

$q^2=\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}$

$q=\sqrt{\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}}$

$q=2.82\times10^{-6}\ C$

Hence, The magnitude of each charge is $2.82\times10^{-6}\ C$

You might be interested in

Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 350 newtons stretches a sprin

Bingel [31]

Answer:

Explanation:

350 N force stretches the spring by 30 cm

spring constant K = 350 / 0.30 = (350 / 0.3) N / m

To calculate work done by a spring force we proceed as follows

spring force when the spring is stretched by x = Kx

This force is variable so work done by it can be calculated by integration

Work done by it in stretching from x₁ to x₂

W = ∫ F dx

= ∫ Kx dx with limit from x₁ to x ₂

= 1/2 K ( x₂² - x₁² )

Putting the given values of x₁ = 0.50 m , x₂ = 0.8 m

Work done

= 1/2 x (350 / 0.3)x ( 0.80² - 0.50² )

= 227.50 J

5 0

3 years ago

An upward force act on a proton as it moves with a speed of 2.0 x 10^5 meters/seconds through a magnetic field of 8.5 x 10^2

Gekata [30.6K]

Force on a moving charge is given by formula

$\vec F = q(\vec v \times \vec B)$

here we know that this force will be maximum when velocity is perpendicular to magnetic field

$\vec F = qvB$

here we know that

$v = 2.0 \times 10^5 m/s$

$q = 1.6 \times 10^{-19} C$

$B = 8.5 \times 10^2 T$

now we have

$F = (1.6 \times 10^{-19})(2 \times 10^5)(8.5 \times 10^2)$

$F = 2.72 \times 10^{-11} N$

7 0

3 years ago

a fan is rotating clockwise and its acceleration has a positive sign. is the angular velocity of the fan speeding up, slowing do

balu736 [363]

Answer:

The angular velocity is slowing down.

Explanation:

By convention, if a rigid body is rotating clockwise, the angular velocity is negative.
If the angular acceleration has a positive sign, since the angular acceleration and the angular velocity have opposite signs, this means that the angular velocity is slowing down.

5 0

2 years ago

Yub87654d HELP there is this person trying to kid,.,nap,./, kids so if you see an account called Hernyana report there questions

Morgarella [4.7K]

Answer:

what???

Explanation:

7 0

3 years ago

If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.

grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution. By definition is defined as:

$\theta = 1.22\frac{\lambda}{d}$

Where,

$\lambda$ = Wavelength

d = Width of the slit

$\theta$ = Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

$r = l\theta$

Relacing $\theta$

$r = l(\frac{1.22\lambda}{d})$

$r = 1.22\frac{l\lambda}{d}$

Replacing with our values we have that,

$r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})$

$r = 3680.91m$

Therefore the diameter of the beam on the moon is

$d = 2r$

$d = 2 * (3690.91)$

$d = 7361.8285m$

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0

3 years ago