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kirza4 [7]
3 years ago
14

Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value o

f consecutive lines is constant. Clear the equipotential lines using the Clear button on the voltage tool. Place the first equipotential line 1 m\rm m away from the charge. It should have a value of roughly 9 V\rm V. Now, produce several additional equipotential lines, increasing and decreasing by an interval of 3 V\rm V (e.g., one with 12 V\rm V, one with 15 V\rm V, and one with 6 V\rm V). Don�t worry about getting these exact values. You can be off by a few tenths of a volt.Which statement best describes the distribution of the equipotential lines?1-The equipotential lines are closer together in regions where the electric field is weaker.2-The equipotential lines are closer together in regions where the electric field is stronger.3-The equipotential lines are equally spaced. The distance between each line is the same for all adjacent lines.
Physics
1 answer:
Anna35 [415]3 years ago
3 0

Answer:

the correct one is 2. the equipotential lines must be closer together where the field has more intensity

Explanation:

The equipotential line concept is a line or surface where a test charge can move without doing work, therefore the potential in this line is constant and they are perpendicular to the electric field lines.

In this exercise we have a charge and a series of equipotential lines, if this is a point charge the lines are circles around the charge, where the potential is given by

           V = k q / r

also the electric field and the electuary potential are related

           E =  - \frac{dV}{dr}

therefore the equipotential lines must be closer together where the field has more intensity

When checking the answers, the correct one is 2

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Block A, with a mass of 4.0 kg, is moving with a speed of 2.0 m/s while block B , with a mass of 8.0 kg, is moving in the opposi
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Consider velocity to the right as positive.

First mass:
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3 years ago
Students run an experiment to determine the rotational inertia of a large spherically shaped object around its center. Through e
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<h2>ANSWER:</h2>
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Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun
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Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and  i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

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i₃ = -(0.40 A - 0.65 A)

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(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

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