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andrezito [222]
3 years ago
5

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

roton and an electron are situated 911 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed.
Physics
1 answer:
coldgirl [10]3 years ago
4 0

Answer:

Explanation:

charge, q = 1.6 x 10^-19 C

distance, r = 911 nm = 911 x 10^-9 m

The Coulomb's force is given by

F=\frac{Kq_{1}q_{2}}{r^{2}}

F=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{\left (911\times 10^{-9}  \right )^2}

F = 2.78 x 10^-16 N

The force between the electron and the proton is 2.78 x 10^-16 N.

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The force between a pair of .005 charges is 750 N. What is the distance between them?
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Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?

Answer:

17.32 m

Explanation:

From coulomb's Law,

F = kqq'/r²........................... Equation 1

Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.

make r the subject of the equation above

r = √(kqq'/F)..................... Equation 2

From the question,

Given: q = q' = 0.005 C, F = 750 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute these values into equation 2

r = √(9.0×10⁹×0.005×0.005/750)

r = √(300)

r = 17.32 m.

Hence the distance between the pair of charges = 17.32 m

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Why does light change its velocity (or slows down) when it encounters a glass lens?​
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A charge of 1.0 × 10-6 μC is located inside a sphere, 1.25 cm from its center. What is the electric flux through the sphere due
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Answer:

\Phi_E=0.11\frac{N\cdot m^2}{C}

Explanation:

According to Gauss's Law, the electric flux of a charged sphere is the electric field multiplied by the area of ​​the spherical surface:

\Phi_E=EA\\\Phi_E=E(4\pi R^2)\\\Phi_E=\frac{q}{4\pi \epsilon_0 R^2}(4\pi R^2)\\\Phi_E=\frac{q}{\epsilon_0}

This is identical to the electric flux of a point charge located in the center of the sphere.

\Phi_E=\frac{1*10^{-12}C}{8.85*10^{-12} \frac{C^2}{N\cdot m^2}}\\\Phi_E=0.11\frac{N\cdot m^2}{C}

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