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dusya [7]
3 years ago
9

A current runs from the left to the right in a long, straight wire. How does the magnetic field at point X compare with the magn

etic field at point Y?
A. The magnetic field at point X points into the page, and the magnetic field at point Y points into the page.

B. The magnetic field at point X points out of the page, and the magnetic field at point Y points out of the page.

C. The magnetic field at point X points out of the page, and the magnetic field at point Y points into the page.

D. The magnetic field at point X points into the page, and the magnetic field at point Y points out of the page.
Physics
1 answer:
vitfil [10]3 years ago
7 0
I think the answer is <span>D. The magnetic field at point X points into the page, and the magnetic field at point Y points out of the page.</span>
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A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.
Volgvan

A) 16.1 N

The magnitude of the electric force between the corks is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C is the magnitude of the charge on the first cork

q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C is the magnitude of the charge of the second cork

r = 0.12 m is the separation between the two corks

Substituting numbers into the formula, we find

F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N

B) Attractive

According to Coulomb's law, the direction of the electric force between two charged objects depends on the sign of the charge of the two objects.

In particular, we have:

- if the two objects have charges with same sign (e.g. positive-positive or negative-negative), the force is repulsive

- if the two objects have charges with opposite sign (e.g. positive-negative), the force is attractive

In this problem, we have

Cork 1 has a positive charge

Cork 2 has a negative charge

So, the force between them is attractive.

C) 2.69\cdot 10^{13}

The net charge of the negative cork is

q_2 = -4.3 \cdot 10^{-6}C

We know that the charge of a single electron is

e=-1.6\cdot 10^{-19}C

The net charge on the negative cork is due to the presence of N excess electrons, so we can write

q_2 = Ne

and solving for N, we find the number of excess electrons:

N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}

D) 3.75\cdot 10^{13}

The net charge on the positive cork is

q_1 = +6.0\cdot 10^{-6}C

We know that the charge of a single electron is

e=-1.6\cdot 10^{-19}C

The net charge on the positive cork is due to the "absence" of N excess electrons, so we can write

q_1 = -Ne

and solving for N, we find the number of electrons lost by the cork:

N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}

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3 years ago
Who wants to add me on Insta
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Answer:

yeah meeeeee will add you

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What happens when the object is placed at F? Explain<br> your answer.
34kurt

Answer:

Sample Response: No image will be formed because the rays will not converge to or diverge from a common point.

Explanation:

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What could you do to increase the electric potential energy between two positively charged particles by a factor of 16?
ivann1987 [24]
Electric~potential ~energy~U_{E}=K_{e}.\dfrac{Qq}{R}\\so~ U_{E} \propto \dfrac{1}{R}
so to increase Ue by 16 times you gotta reduce R(distence) by 16 times
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Answer:

HI!!!

Explanation:

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