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Paraphin [41]
2 years ago
10

A small metal ball is given a negative charge, then brought near to end a of the rod (figure 1). What happens to end a of the ro

d when the ball approaches it closely this first time?.
Physics
1 answer:
erma4kov [3.2K]2 years ago
7 0

What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.

<h3>Electrostatics</h3>

I have attached the image of the rod.

We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.

This means that their fields will cancel.

Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.

This also applies to a strong conducting rod and therefore it is strongly attracted.

Read more about Electrostatics at; brainly.com/question/18108470

You might be interested in
Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has
faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

  • Charge on first charged particle, q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.
  • Charge on the second charged particle, q_2=3.80\ nC=3.80\times 10^{-9}\ C.
  • Position of the first charge = (x_1=0.00\ m,\ y_1=0.600\ m).
  • Position of the second charge = (x_2=1.50\ m,\ y_2=0.650\ m).

The electric field at a point due to a charge q at a point r distance away is given by

\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.

where,

  • k = Coulomb's constant, having value \rm 8.99\times 10^9\ Nm^2/C^2.
  • \vec r = position vector of the point where the electric field is to be found with respect to the position of the charge q.
  • \hat r = unit vector along \vec r.

The electric field at the origin due to first charge is given by

\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.

\vec r_1 is the position vector of the origin with respect to the position of the first charge.

Assuming, \hat i,\ \hat j are the units vectors along x and y axes respectively.

\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.

Using these values,

\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.

The electric field at the origin due to the second charge is given by

\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.

\vec r_2 is the position vector of the origin with respect to the position of the second charge.

\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.

Using these values,

\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.

The net electric field at the origin due to both the charges is given by

\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0
3 years ago
What describes a sound wave as it travels through a medium
Softa [21]
Sound waves in air (and any fluid medium) are longitudinal waves because particles of the medium through which the sound is transported vibrate parallel to the direction that the sound wave moves.
5 0
3 years ago
A 65-kg person stands on a scale in a moving elevator while holding a 5.0 kg mass suspended from a massless spring with spring c
disa [49]

Answer:

The question is incomplete. However, I believe, it is asking for the acceleration of the elevator. This is 3.16 m/s².

Explanation:

By Hooke's law, F = ke

F is the force on a spring, k is the spring constant and e is the extension or compression.

From the question,

F = (1.08\text{ kN/m}) \times (6.0 \times 10^{-2}\text{ m}) = 64.8 \text{ N}

This is the force on the mass suspended on the spring. Its acceleration, a, is given by

F = ma

a = \dfrac{F}{m}

a = \dfrac{64.8 \text{ N}}{5\text{ kg}} = 12.96\text{ m/s}^2

This acceleration is more than the acceleration due to gravity, g = 9.8 m/s². Hence the elevator must be moving up with an acceleration of

12.96 - 9.8 m/s² = 3.16 m/s²

7 0
3 years ago
Anyone know what the answer is?..
Licemer1 [7]

Answer:

sometimes harmful and sometimes beneficial

8 0
4 years ago
Micah knows that a car had a change in velocity of 15 m/s.To determine acceleration ,Micah also needs the blank of the total tri
Fittoniya [83]

Answer: To determine acceleration ,Micah also needs the Time of the total trip in seconds.

Explanation:

Acceleration can be defined as rate of change of velocity.

a = \frac{dv}{dt}

for calculating acceleration, initial and final velocity are required in meter per second and the total time of the trip in seconds. Then acceleration is measured in meter per second square.

Thus, Micah knows that a car had a change in velocity of 15 m/s.To determine acceleration ,Micah also needs the <u>Time</u> of the total trip in seconds.

7 0
4 years ago
Read 2 more answers
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