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lesya692 [45]
2 years ago
8

In which case would electrical potential energy be built up and stored in the electric field? a) A positive charge is moved towa

rd a negative charge. b) A positive charge is moved toward a positive charge. c) A negative charge is moved away from a negative charge. d) none of the above
Physics
1 answer:
mr_godi [17]2 years ago
5 0

Answer:

The correct option is B

Explanation:

Although, it is common knowledge that in an electric field, unlike charges attract and like charges repel. However, to build up an electric potential, a positive charge is brought close to another positive charge - this causes work done to be changed to electric potential energy and stored in the electric field.

It should however be noted that when a negative charge is moved away from a positive charge, the negative charge gains electric potential energy.

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What is the most significant challenge for organisms that live in estuaries?
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"Changing water salinity" is the most significant challenge for organisms that live in estuaries.

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8 0
3 years ago
What is E(r)E(r)E(r), the radial component of the electric field between the rod and cylindrical shell as a function of the dist
andriy [413]

Answer:

E(r) = λ/2πrε0

Explanation:

If we consider an infinitely long line of charge with the charge per unit length being λ, we can take advantage of the cylindrical symmetry of this situation.

By symmetry, i mean that the electric fields all point radially away from the line of charge and thus there is no component parallel to the line of charge.

Niw, let's use a cylinder (with an arbitrary radius (r) and length (l)) centred on the line of charge as our Gaussian surface.

Doing that will mean that the electric field would be perpendicular to the curved surface of the cylinder. Therefore, the angle between the electric field and area vector is equal to zero and cos θ = cos 0 = 1

Now, the top and bottom surfaces of the cylinder will lie parallel to the electric field. Therefore, the angle between the area vector and the electric field would be 90° and cos θ = cos 90 = 0

Now, we know that according to Gauss Law,

Electric Flux, Φ = E•dA

Thus,

Total Φ = Φ_curved + Φ_top + Φ_bottom

Thus,

Φ = ∫E•dA cos 0 + ∫E•dA cos 90° + ∫E•dA cos 90°

We now have ;

Φ = ∫E . dA × 1

Since we are dealing with the radial component, the curved surface would be equidistant from the line of charge and the electric field in the surface will be the same magnitude throughout.

Thus,

Φ = ∫E•dA = E∫dA = E•2πrl

The net charge enclosed by the surface is given by:

q_net = λl

So using gauss theorem, we have;

Φ = E•2πrl = q_net/εo = λl/εo

E•2πrl = λl/ε0

Making E the subject, we obtain ;

E = λ/2πrε0

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Hope this helped.

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