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2 years ago

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The Coulomb force between two charges q1 and q2 at separation r in the air is 10N. If half of the separation is filled with medi

um of dielectric constant 7, what will be the value of new coulomb force?

1 answer:

adoni [48]2 years ago

5 0

Answer:

The value of new coulomb force is 1.43 N.

Explanation:

Given;

Coulomb's force in vacuum (air), $F_v$ = 10 N

dielectric constant, K = 7

The Coulomb's force between two charges separated by a distance r in a vacuum is given as;

$F_v = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2}$

The Coulomb's force between two charges separated by a distance r in a medium with dielectric constant is given as

$F_m = \frac{1}{4\pi K\epsilon_0} \frac{q_1q_2}{r^2}$

Take the ratio of the two forces;

$\frac{F_v}{F_m} = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2} \ \times \ \frac{4\pi K\epsilon_0 r^2}{q_1q_2} = K\\\\\frac{F_v}{F_m} = K\\\\\frac{10}{F_m} = 7\\\\F_m = \frac{10}{7} \\\\F_m = 1.43 \ N$

Therefore, the value of new coulomb force is 1.43 N.

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What is the prefix for oxygen in As2O5? tri- di- mono- penta-

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The prefix for oxygen in As2O5 is PENTA.
Penta is used to show that the number of oxygen atoms present in the compound is five. Penta means five, while mono mean one, di means two, tri means three and tetra mans four and so on. The chemical name for the given compound is Arsenic pentoxide.

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3 years ago

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Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet

sveta [45]

Answer:

The level of the root beer is dropping at a rate of 0.08603 cm/s.

Explanation:

The volume of the cone is :

$V=\frac {1}{3}\times \pi\times r^2\times h$

Where, V is the volume of the cone

r is the radius of the cone

h is the height of the cone

The ratio of the radius and the height remains constant in overall the cone.

Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm

h = 13 cm

r / h = 5 / 13

r = {5 / 13} h

$V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h$

$V=\frac {550}{3549}\times h^3$

Also differentiating the expression of volume w.r.t. time as:

$\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}$

Given: $\frac {dV}{dt}$ = -4 cm³/sec (negative sign to show leaving)

h = 10 cm

So,

$-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}$

$\frac{55000}{1183}\times \frac {dh}{dt}=-4$

$\frac {dh}{dt}=-0.08603\ cm/s$

<u>The level of the root beer is dropping at a rate of 0.08603 cm/s.</u>

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2 years ago

If you stood on a planet with four times the mass of Earth, and twice Earth's radius, how much would you weigh?

nikdorinn [45]

Answer:

1/4 times your earth's weight

Explanation:

assuming the Mass of earth = M

Radius of earth = R

∴ the mass of the planet= 4M

the radius of the planet = 4R

gravitational force of earth is given as = $\frac{GM}{R^{2} }$

where G is the gravitational constant

Gravitational force of the planet = $\frac{G4M}{(4R)^{2} }$

= $\frac{G4M}{16R^{2} }$

= $\frac{GM}{4R^{2} }$

recall, gravitational force of earth is given as = $\frac{GM}{R^{2} }$

∴Gravitational force of planet = 1/4 times the gravitational force of the earth

you would weigh 1/4 times your earth's weight

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3 years ago

An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of

Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, $E=390\ keV=390\times 10^3\ eV$

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2meE}}$

$\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}$

$\lambda=1.96\times 10^{-12}\ m$

(b) According to Bragg's law,

$n\lambda=2d\ sin\theta$

n = 1

For nickel, $d=0.092\times 10^{-9}\ m$

$\theta=sin^{-1}(\dfrac{\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})$

$\theta=0.010^{\circ}$

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

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3 years ago

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.2m/s and the hatchback ca

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Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. $s_{o}$ = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.

Truck:

$v_{i}$ = 33.2 m/s, a = 0 (since the velocity is constant), $s_{o}$ = 0

Using $s =s_{o}+v_{i}t+1/2at^{2}$

s = 33.2t .......... eq (1)

Hatchback:

$a=5m/s^{2}$ , $v_{i}$ = 0 m/s (since initial velocity is zero), $s_{o}$ = 0

Using $s =s_{o}+v_{i}t+1/2at^{2}$

putting in the data we will get

$s=(1/2)(5)t^{2}$

now putting 's' value from eq (1)

$2.5t^{2}-33.2t=0$

which will give,

t = 13.28 s

so both vehicles will meet up gain after 13.28 sec.

putting t = 13.28 in eq (1) will give

s = 440.896 m

So, both vehicles will meet up again at 440.896 m.

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3 years ago

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