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
slava [35]
3 years ago
7

A long line carrying a uniform linear charge density+50.0μC/m runs parallel to and 10.0cm from the surface of a large, flat plas

tic sheet that has a uniform surface charge density of −100μC/m2on one side.Part AFind the location of all points where an α particle would feel no force due to this arrangement of charged objects.L=_____________ m from the line.Part BChoose an appropriate location of these points at a distance calculated in part A.o above the lineo between the line and the sheet.
Physics
1 answer:
IRISSAK [1]3 years ago
6 0

Answer:

The location of the points is at the distance  above the line charge Option A.

Explanation:

Part A

The electric field due to a line of charge is  expressed as :

Eₐ = λ /2πε₀L  …… (1)

while  the electric field due to a non-conducting elastic sheet is expressed as:

Еₙ =   λ /2ε₀  …… (2)

In order for the  particle to experience no force, the electric field at that point is zero.

Equating  (1) and (2) together, we have:.

λ /2πε₀L = σ /2ε₀  

Rearranging the above expression, we have:

L =  λ /πσ

Substituting 50.0 μC/m  for λ, 3.14 for  π  , and 100 μC/ m² for σ, we have:

L=  50.0 μC/m / (3.14)* 100 μC/ m²

L = 0.159m

From the L value obtained above, the location of all points where an  particle would experience no force due to the arrangement of charged objects is L 0.159m  from the line.

Part B  

The electric field is zero at all the points above the line with the distance 0.159m  from the line.

Thus, the location of the points is at the distance  above the line charge Option A.

You might be interested in
When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is referred to as
Dafna11 [192]
When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is called "load triangle". Because of a motorcycle's size and weight<span> and the fact that it has only two wheels, how to carry extra load is very important. One has to make sure that they are keeping the weight low and close to the middle of the motorcycle and keep the load evenly from side to side. Heavier items should be in the "load triangle".</span><span> </span>
3 0
3 years ago
Which statement is true of gravity?
o-na [289]

Answer:

B is the answers for the question

3 0
2 years ago
Before the fission process takes place, lead-207 is bombarded with neutrons, it can change into
Oduvanchick [21]

Lead-207 can change into Lead-208.

When Lead-207 is bombarded with neutrons, the atom acquires the neutron. Adding a neutron only changes the mass number of the atom. This does not involve change in the identity of the atom. So, Lead-207 can change into Lead-208 without change in its chemical properties.

8 0
3 years ago
Read 2 more answers
You are driving your 1700 kg car at 21 m/s down a hill with a 5.0∘ slope when a deer suddenly jumps out onto the roadway. You sl
podryga [215]

Answer:

d = 27.7 m

Explanation:

Here the car is driving on the inclined plane

So here we can say that work done by the gravity and work done by friction is equal to change in kinetic energy of the system

So here we can write it as

mgsin\theta \times d - F_f \times d = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

now we have

m = 1700 kg

v_f = 0

v_i = 21 m/s

F_f = 1.5 \times 10^4 N

(1700)(9.81)sin5 (d) - (1.5\times 10^4)d = 0 - \frac{1}{2}(1700)(21^2)

1453.5 d - (1.5\times 10^4)d = -374850

d = 27.7 m

6 0
3 years ago
The compressed-air tank ab has a 250-mm outside diameter and an 8-mm wall thickness. it is fitted with a collar by which a 40-kn
valentinak56 [21]
<span>Assume: neglect of the collar dimensions. Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa Ď„=(S*Q)/(I*b)=(40*〖10〗^3*Ď€(〖0.125〗^2-〖0.117〗^2 )*121*〖10〗^(-3))/(Ď€/2 (〖0.125〗^4-〖0.117〗^4 )*8*〖10〗^(-3) )=41.277 MPa @ Point K: Ď_z=(+M*c)/I=(40*0.6*121*〖10〗^(-3))/(8.914*〖10〗^(-5) )=32.6 MPa Using Mohr Circle: Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 ) Ď_max=104.2 MPa, Ď„_max=45.62 MPa</span>
3 0
3 years ago
Other questions:
  • A gas is __________ and assumes __________ of its container whereas a liquid is __________ and assumes __________ of its contain
    10·1 answer
  • Anna studies the color of her bedroom wall using different lights. In sunlight, she sees that the wall is white. Anna darkens th
    14·2 answers
  • A softball is hit over a third baseman's head with some speed v0 at an angle θ above the horizontal. Immediately after the ball
    5·1 answer
  • Critical mass depends on ___. Check all that apply.
    8·1 answer
  • A quasar is a distant celestial body in space. Investigators use a special telescope and determine that a certain quasar was giv
    7·1 answer
  • HELP THIS IS DUE IN 5 MINUTES!!!!!!!!!!!! WILL GIVE BRAINLIEST
    8·1 answer
  • How does temperature affect the volume and pressure of a gas?
    10·1 answer
  • How can the pilot determine, for an ILS runway equipped with MALSR, that there may be a penetration of the obstacle identificati
    12·1 answer
  • What is the job of the front center position? a. Set the ball b. Receive the ball c. Serve the ball
    7·2 answers
  • Describe what the sun would look like from earth if the entire photosphere were the same temperature as a sunspot?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!