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3 years ago

15

Threepointchargeshaveequalmagnitudesandarelocatedonthesameline. The separation d between A and B is the same as the separation b

etween B and C. One of the charges is positive and two are negative, as the drawing shows. Consider the net electrostatic force that each charge experiences due to the other two charges. Rank the net forces in descending order (greatest first) according to magnitude. (a) A, B, C (b) B, C, A (c) A, C, B (d) C, A, B (e) B, A, C

2 answers:

Zinaida [17]3 years ago

7 0

Answer:

the descending order of the forces is

B A C

Therefore the correct result is e

Explanation:

To answer this exercise we must use that the forces are vector magnitudes and the value is given by Coulomb's law

F = k q₁q₂ / x²

Consider the following scheme

A B C

-q + q + q

x x

The forces are

F_AB = k q² / x²

F_AC = k q² / 4x²

Fo = k q² / x²

F_AB = Fo

F_AC = ¼ Fo

F_BC= F_AB

The forces on the charge A are

F_A = F_AB + F_AC

F_A = Fo + ¼ Fo

F_A= 5/4 Fo

The forces on the charge B

F_B = -F_AB - F_BC

F_B = - Fo - Fo

F_B = -2 Fo

The forces on the load C

F_C = F_BC –F_AB

F_C = Fo - ¼ Fo

F_C = ¾ Fo

For this distribution of charges the descending order of the forces is

B A C

Therefore the correct result is e

If the charge distribution has the scheme

A B C

+ q -q +q

x x

Force on A

F_A = FA_B - F_BC

F_A = Fo - ¼ Fo = ¾ Fo

Force on B

F_B = F_AB –F_BC

F_B = 0

Force on C

F_C = F_AC - F_BC

F_C = ¼ Fo - Fo

F_C = - ¾ Fo

For this distribution of charges the descending order of the forces is

A B C

result is a

Molodets [167]3 years ago

3 0

Answer: B, C, A

Explanation:

Electrostatic force F = kq²/d²

Let the force on A be Fa, and B be Fb and C be Fc.

For Fa, distance between A and B is d, and distance between A and C is 2d

For Fb, distance between B and A is d, and distance between B and C is d

For Fc, distance between C and B is d, and distance between C and A is 2d

Also, the charge of each charge affects the sign of the force.

Check the attachment below

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Answer:

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The illustration of the problem is shown the attached image.

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3 years ago

Plz help ASAP I'll mark as brainliest

gogolik [260]

Hi there!

1.

Hooke's law states that:

F = -kx

k = Spring constant (N/m)

x = DISPLACEMENT from equilibrium (m)

Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.

2.

The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)

3.

The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.

4.

We can begin by looking at the given graph.

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5.1.

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5.2.

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The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:

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