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NARA [144]
3 years ago
5

Fill in the blanks. The electrostatic force between two objects is proportional to the ____________________ of the distance ____

________________.
Physics
2 answers:
Shkiper50 [21]3 years ago
6 0
Write an equation to calculate the force between two objects if the product of their charges is 10.0 × 10-4 C. (Note: Use the variable R for the distance between the charges.)

F = 900 ÷_________
Luba_88 [7]3 years ago
6 0

Answer:

The electrostatic force between two objects is proportional to the inverse of the square of the distance between the centres of the two objects

Explanation:

According to Coulomb's law of electrostatic attraction, the electrostatic force F between two charges or objects is directly proportional to the product of the absolute values of the charges or (charges on the objects) and inversely proportional to the square of their distance of separation.

Given that q_1 and q_2 are the charges or the charges on the objects, we can break down Coulomb's statement mathematically as follows;

F\alpha q_1q_2.............(1)  F is proportional to the product of the charges.

F\alpha \frac{1}{r^2}.......................(2) F is proportional to the inverse of the square of their distance r of separation.

This is an inverse-square relationship.

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A book is sitting on a desk. What best describes the normal force acting on the book?
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Which of the following expressions gives the ratio of the energy density of the magnetic field to that of the electric field jus
miss Akunina [59]

Answer:

(d) \ \ \frac{\mu_o}{\epsilon_o} (\frac{L}{2\pi r*R} )^2

Explanation:

Energy density in magnetic field is given as;

U_B = \frac{1}{2 \mu_o} B^2

where;

B is the magnetic field strength

Energy density of electric field

U_E = \frac{1}{2}\epsilon E^2

where;

E is electric field strength

Take the ratio of the two fields energy density

\frac{U_B}{U_E} = \frac{1}{2\mu_o} B^2 / \frac{1}{2}\epsilon E^2\\\\\frac{U_B}{U_E} = \frac{B^2}{2\mu_o}  *\frac{2}{\epsilon E^2} \\\\\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B^2}{E^2})

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B}{E})^2

But, Electric field potential, V = E x L = IR (I is current and R is resistance)

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{E*L})^2

Now replace E x L with IR

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{IR})^2

Also, B = μ₀I / 2πr, substitute this value in the above equation

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{\mu_oI*L}{2\pi r* IR})^2

cancel out the current "I" and factor out μ₀

\frac{U_B}{U_E} = \frac{\mu_o^2}{\mu_o \epsilon} (\frac{L}{2\pi r* R})^2

Finally, the equation becomes;

\frac{U_B}{U_E} = \frac{\mu_o}{\epsilon} (\frac{L}{2\pi r*R })^2

Therefore, the correct option is (d) μ₀/ϵ₀ (L /R 2πr)²

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