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Mashcka [7]
3 years ago
8

Given the mathematical representation of Coulomb’s Law, , where , describe in words the relationship among electric force, charg

e, and distance.
Physics
1 answer:
Svetach [21]3 years ago
5 0

Answer:

F_c = \frac{KQ_1Q_2}{R^2}

Explanation:

Coulomb’s law is given as;

F_c = \frac{KQ_1Q_2}{R^2}

where;

F is the electric force between q₁ and q₂

k is the coulomb’s constant  = 9 x 10⁹ Nm²/C²

r is the distance between the two charges

q₁ and q₂ are the two charges

Therefore, the coulomb’s law is given as F_c = \frac{KQ_1Q_2}{R^2}

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If the depth of water in a well is 10 m, what is the pressure exerted by it on the
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3 years ago
A coil 3.95 cm radius, containing 520 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×1
Pie

Answer:

(a) E= 3.36×10−2 V +( 3.30×10−4 V/s3 )t3

(b) I=0.0085\ A

Explanation:

Given:

  • radius if the coil, r=0.0395\ m
  • no. of turns in the coil, n=520
  • variation of the magnetic field in the coil, B=(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4
  • resistor connected to the coil, R=560\ \Omega

(a)

we know, according to Faraday's Law:

emf=n.\frac{d\phi}{dt}

where:

d \phi= change in associated magnetic flux

\phi= B.A

where:

A= area enclosed by the coil

Here

A=\pi.r^2

A=\pi\times 0.0395^2

A=0.0049\ m^2

\therefore \phi=((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049

So, emf:

emf= 520\times \frac{d}{dt} [((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049]

emf= 520\times 0.0049\times \frac{d}{dt} [(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)]

emf= 2.548\times [0.012+(13.8\times 10^{-5})t^3)]

emf= 0.0306+3.516\times 10^{-4}\ t^3

(b)

Given:

t_0=5.25\ s

Now, emf at given time:

emf=4.7755\times 10^{-2}\ V

∴Current

I=\frac{emf}{R}

I=\frac{4.7755\times 10^{-2}}{560}

I=8.5\times 10^{-5} A

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