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patriot [66]
3 years ago
14

Imagine a universe in which, like in ours, there are two kinds of charges (positive and negative), with the like charges repelli

ng and unlike charges attracting each other, but with the electric force somewhat different from Coulomb's Law. Instead of varying with inverse distance squared, the force between charges in this imagined world would vary with the inverse distance cubed. In such a universe, would the static charges still always be distributed on the surfaces of conductors?
Physics
1 answer:
GuDViN [60]3 years ago
5 0

Answer:

the static charge is not always distributed on the surface of the conductor, there are also charges in the volume but of lesser magnitude

Explanation:

In this hypothetical system the electric force is of type

       F = k' \frac{q_1 q_2 }{r^2}

in this case the force decays to zero much faster,

if we call Fo the force of Coulomb's law

         F₀ = k \frac{q_1 q_2 }{r^2}

assuming the constant k is the same

the relationship between the two forces is

        F / F₀ = 1 / r

        F = F₀ / r

when analyzing this expression the force decays much faster to zero.

In an electric conductor, charges of the same sign may not feel any repulsive force from other charges that are at a medium distance, so there is a probability that some charges are distributed in the volume of the material, this does not happen with coulomb's law

Consequently, the static charge is not always distributed on the surface of the conductor, there are also charges in the volume but of lesser magnitude

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8 0
2 years ago
An electric heater used to boil small amounts of water consists of a 15-Ω coil that is immersed directly in the water. It operat
Allisa [31]

Answer:

t = 1120 seconds

Explanation:

Law of Conservation of Energy: It states that energy can neither be created or destroyed but it can be transformed from one form to another

The heater converts electric energy to heat energy.

According to the law of conservation of Energy,

Heat supplied by the heater = heat gained by the water.

Heat supplied by the heater = (V²/R)t, Where V = voltage, R = Resistance, t = Time.

Heat gained by the water = cm(θ₂ -θ₁)

Where c = specific heat capacity of water, m = mass of water, θ₂= final temperature of water, θ₁= initial temperature of water.

∴ (V²/R)t =  cm(θ₂ -θ₁)

Making  t the subject of the equation,

t =  cm(θ₂ -θ₁) ×R/V²

Where c= 4200 Jkg/K, m=0.80 kg, R = 15 Ω, V = 60 V, θ₁= 20°C, θ₂ = the normal boiling point of water = 100°C

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∴  The time required by the heater to raise the temperature of water to normal boiling point is 1120 seconds.

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3 years ago
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