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faltersainse [42]
1 year ago
8

Points a and b are nearby a proton in otherwise empty space. the potential is chosen to be zero very far away from the proton. t

he potential at a is half as much as the potential at b. compared to the electric field energy density at b, the electric field energy density at a is __
Physics
1 answer:
sertanlavr [38]1 year ago
3 0

Compared to the electric field energy density at b, the electric field energy density at a is large by a factor of 16.

<h3>What is potential energy?</h3>

The energy of a particle by virtue of its position.

The potential at B = Vb =V

Vb = 1/4π∈₀ x e/r₁ = ke/r₁

V =ke/r₁

and r₁ = ke /V

The potential at A = Va = 2V

2V = ke/r₂

r₂ = ke/2V

The electric field at point B. Eb = ke/r₁²

Substitute the value of r₁, we get

Eb = V²/ke

The electric field at point A , Ea = ke /r₂²

Substitute the value of r₂, we get

Ea = 4V²/ke


The electric field energy density at point A and B,

Ua  = 1/2 ∈Ea²

Ub = 1/2 ∈Eb²

electric field energy density is proportional to the square of electric field at the particular point.

Ua/Ub =(Ea/Eb)²

Substitute the values, we get

Ua/Ub = 16

Thus, compared to the electric field energy density at b, the electric field energy density at a is large by a factor of 16.

Learn more about potential energy.

brainly.com/question/24284560

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