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goldfiish [28.3K]
2 years ago
12

What is the current in a 160V circuit if the resistance is 2Ω? V= I= R=

Physics
1 answer:
Alex787 [66]2 years ago
4 0

Explanation:

v = IR

v= 160 R = 2

<u>160</u> = <u>2I</u>

2 2

I = 80A

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The magnitude of electrical force between a pair of charged particles is ____ as much when the particles are moved half as far a
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The magnitude of the electrical force between a pair of charged particles is 4 Times as much when the particles are moved half as far apart.

This can be easily understood by Columb's law,

F_{new} = \frac{kQ_{1}Q_{2}}{r^{2}}

which state's that the amount of electrical force experienced by two charged particles is inversely proportional to the square of the distance between them.

∴ \frac{F_{new} }{F_{old} } = \frac{Distance_{new}^{2}  }{Distance_{old}^{2}  }

Now, we know the new distance is half the original distance,

F_{new} = \frac{kQ_{1}Q_{2}}{\frac{r}{2}^{2} } \\F_{new} = 4\frac{kQ_{1}Q_{2}}{r^{2}}

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The electrical force of attraction or electrostatic force of attraction between two charged particles refers to the amount of attractive or repulsive force that exists between the two charges. This can be calculated by Columb's Law.

A charged particle in physics is a particle that has an electric charge. It might be an ion, such as a molecule or atom having an excess or shortage of electrons in comparison to protons. The same charge is thought to be shared by an electron, a proton, or another primary particle.

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Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
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Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

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