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Schach [20]
1 year ago
15

The potential at points a and b are: va = -2 v and vb = 6 v. when a 5 c charge is moved from a to b, the change in potential ene

rgy of this system is answer j.
Physics
1 answer:
zhenek [66]1 year ago
6 0

The change in potential energy of this system is = 40 J.

<h3>What is the potential difference?</h3>

The potential difference, often known as voltage, is equal to the amount of current times the resistance. One Joule, or one Volt, of energy is required for one Coulomb of charge to flow from one place in a circuit to another.

<h3>What is the formula for potential difference?</h3>

V=Uq The change in potential energy of a charge q transported from point A to point B, divided by the charge, is what is used to determine the electric potential difference between points A and B, or VBVA. The joules per coulomb unit of potential difference is called the volt (V).

This system's potential energy changed by 40 J.

Energy = charge * potential difference

=> Energy = (8v)*5

=> Energy = 40 J

The change in potential energy of this system is = 40 J.

To know more about Potential Difference visit:

brainly.com/question/12198573

#SPJ4

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3 0
2 years ago
The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr
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Answer:

The tube should be held vertically and perpendicular to the ground.

Explanation:

Answer: The tube should be held vertically and perpendicular to the ground. The reason is as follows:

Reasoning:

The power lines are parallel to the ground hence, their electric field will be perpendicular to the ground and equipotential surface will be cylindrical.

Hence, if you will put fluorescent tube parallel to the ground then both the ends of the tube will lie on the same equipotential surface and the potential difference will be zero.

So, to maximize the potential the ends of the tube must be on different equipotential surfaces. The surface which is near to the power line has high potential value and the surface which is farther from the line has lower potential value.

hence, to maximize the potential difference, the tube must be placed perpendicular to the ground.

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