1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nikolay [14]
3 years ago
13

A very large sheet of a conductor carries a uniform charge density of on its surfaces. What is the electric field strength 3.00

mm outside the surface of the conductor?
Physics
1 answer:
olga55 [171]3 years ago
5 0

Complete Question

A very large sheet of a conductor carries a uniform charge density of 4.00\  pC/mm^2  on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?

Answer:

The electric field is  E  =  4.5198 *10^{5} \ N/C

Explanation:

From the question we are told that

    The charge density is  \sigma  =  4.00pC  /mm^2 = 4.00 * 10^{-12 } *  10^{6} = 4.00 *10^{-6}C/m

    The position outside the surface is  a  =  3.00 \ mm  =  0.003 \ m

   

Generally the electric field is mathematically represented as

          E  =  \frac{\sigma}{\epsilon _o  }

Where  \epsilon_o is  the permitivity of free space with values  \epsilon _o  = 8.85 *10^{-12}  F/m

substituting values  

           E  =  \frac{4.0*10^{-6}}{8.85 *10^{-12}  }

           E  =  4.5198 *10^{5} \ N/C

You might be interested in
The picture shows an object resting on a balance.
Maslowich

Answer:

4.90kgm^-2

Explanation:

4 0
1 year ago
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
To the nearest tenth, what is the area of the shaded segment when JA=8ft ??
Murrr4er [49]
A=\dfrac{1}{6}\pi 8^2-\dfrac{8^2\sqrt{3}}{4}\approx 5.8\;[ft^2]
Answer: <span>C. 5.8 ft squared</span>
7 0
3 years ago
Read 2 more answers
Me ajudem, Por favor!!!!!!
zzz [600]

Answer:

a)  a=4\,\frac{m}{s^2}

b)  V(t)=4\,t\,+3

c)  V(1)=7 \,\frac{m}{s} \\

d)  Displacement = 22 m

e)  Average speed = 11 m/s

Explanation:

a)

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}

Therefore,  acceleration is a=4\,\frac{m}{s^2}

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

y=m\,x+b\\V(t)=4\,t\,+3

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = \frac{(7+15)\,2}{2} = 22\,\,m

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = \frac{22}{2} = 11\, \,\frac{m}{s}

3 0
3 years ago
A billiard ball moving at 0.5 m/s strikes another identical billiard ball, which is at rest, in an elastic head-on collision on
NikAS [45]

Answer: assuming that the billiard balls are of identical weight the impacted billiard ball will move forward at around 0.5m/s (not considering energy conservation). The ball impacting the 2nd one would stop because most of its Kinetic energy would have been transferred into the not moving ball.

Explanation: hope this helps!

4 0
3 years ago
Other questions:
  • Letters A, B, C, and D represent locations on a bar magnet. Which location has the greatest magnetic force?
    15·2 answers
  • Link between mass and force of gravity?
    7·2 answers
  • Assume: Moving to the right is positive. A(n) 7.7 g object moving to the right at 22 cm/s makes an elastic head-on collision wit
    14·1 answer
  • One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor
    13·1 answer
  • Electronegativity increases when atoms ___
    14·1 answer
  • •A radioactive material A (decay constant λA) decays into a material B (decay constant λB) and then into material C (decay const
    10·1 answer
  • An aircraft flying in a straight
    9·1 answer
  • ¿Qué dice el principio de conservación de la energía?
    5·1 answer
  • I NEED THE RIGHT ANSWER ASAP NO LINKS !!!<br> This is a Science question
    12·2 answers
  • Using the graph above determine the acceleration for the object represented between 6.0s and 8.0s
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!