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3 years ago

The original magnitude of the force on the Q charge was F; what is the magnitude of the force on the Q charge now?

Physics

1 answer:

Otrada [13]3 years ago

3 0

Complete Question:

Two small objects each with a net charge of Q (where Q is a positive number) exert a force of magnitude "F" on each other. We

replace one of the objects with another whose net charge is 4Q. The original magnitude of the force on the Q charge was "F"; what is the magnitude of the force on the Q charge now?

Answer:

4 F₀

Explanation:

Assuming that we can treat to both objects as point charges, we can find the force "F" that one charge exerts upon the other applying Coulomb´s law, as follows:

F₀ = K*Q₀² / r₁₂²

If we replace one of the charges by one with a 4Q₀ charge, the new value of F will be as follows:

F₁ = K*Q₀*4Q₀ / r₁₂² =( K*Q₀² / r₁₂²)* 4 = 4* F₀

This value is reasonable, as the electrostatic force is a linear - type one, so it is possible to use the superposition principle (we can get the force exerted by one charge on another without considering the ones due to another charges)

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3 years ago

We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes

Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

$1.56421\times 10^{-6}\ T$

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = $\frac{d}{2}=\frac{7}{2}=3.5\ cm$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

The intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2$

5% of this energy goes to the visible light so the intensity is

$I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2$

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

$u=\frac{1}{2}\epsilon_0E^2$

Energy density is also given by

$\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m$

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

$B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T$

The amplitude of the magnetic field at this surface is $1.56421\times 10^{-6}\ T$

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3 years ago

Andrea is investigating how increasing the concentration of sugar affects the boiling point of the solution. What would be the b

NNADVOKAT [17]

An increase in the concentration of sugar in a solution leads to the increase in boiling point of the solution.

<h3>What is boiling point?</h3>

Boiling point of a solution is that pressure exerted on a liquid by its surrounding is equal to the vapour pressure of the liquid.

When there is an increase in concentration of solute, more space will be occupied in the fluid leading to an increase in the boiling point of the fluid.

Therefore, An increase in the concentration of sugar in a solution leads to the increase in boiling point of the solution.

Learn more about boiling point here:

brainly.com/question/40140

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