Answer:
The electric field at origin is 3600 N/C
Solution:
As per the question:
Charge density of rod 1, ![\lambda = 1\ nC = 1\times 10^{- 9}\ C](https://tex.z-dn.net/?f=%5Clambda%20%3D%201%5C%20nC%20%3D%201%5Ctimes%2010%5E%7B-%209%7D%5C%20C)
Charge density of rod 2, ![\lambda = - 1\ nC = - 1\times 10^{- 9}\ C](https://tex.z-dn.net/?f=%5Clambda%20%3D%20-%201%5C%20nC%20%3D%20-%201%5Ctimes%2010%5E%7B-%209%7D%5C%20C)
Now,
To calculate the electric field at origin:
We know that the electric field due to a long rod is given by:
![\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%20%7D%7B2%5Cpi%20%5Cepsilon_%7Bo%7D%7BR%7D)
Also,
(1)
where
K = electrostatic constant = ![\frac{1}{4\pi \epsilon_{o} R}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_%7Bo%7D%20R%7D)
R = Distance
= linear charge density
Now,
In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.
At x = - 1 cm = - 0.01 m:
Using eqn (1):
![\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B2%5Ctimes%209%5Ctimes%2010%5E%7B9%7D%5Ctimes%201%5Ctimes%2010%5E%7B-%209%7D%7D%7B0.01%7D%20%3D%201800%5C%20N%2FC)
(towards)
Now, at x = 1 cm = 0.01 m :
Using eqn (1):
![\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C](https://tex.z-dn.net/?f=%5Cvec%7BE%27%7D%20%3D%20%5Cfrac%7B2%5Ctimes%209%5Ctimes%2010%5E%7B9%7D%5Ctimes%20-%201%5Ctimes%2010%5E%7B-%209%7D%7D%7B0.01%7D%20%3D%20-%201800%5C%20N%2FC)
(towards)
Now, the total field at the origin is the sum of both the fields:
![\vec{E_{net}} = 1800 + 1800 = 3600\ N/C](https://tex.z-dn.net/?f=%5Cvec%7BE_%7Bnet%7D%7D%20%3D%201800%20%2B%201800%20%3D%203600%5C%20N%2FC)
Answer:
use their own matter to create nuclear reactions
generate their own energy
are classified primarily by temperature
usually seem bright if they are close to Earth
Explanation:
Explanation:
By the statement, Efficiency of a machine is 60% it is meant that 'The work output is 60% of the work input'. If you apply 100J of energy for a machine, it will be able to life load of 60 J only
The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.
The Coulomb force between the two balls is:
![F= k_e \frac{ q_1 q_2}{r^2}](https://tex.z-dn.net/?f=F%3D%20k_e%20%5Cfrac%7B%20q_1%20q_2%7D%7Br%5E2%7D%20)
where
![k_e=8.99\cdot10^9~N m^2 C^{-2}](https://tex.z-dn.net/?f=k_e%3D8.99%5Ccdot10%5E9~N%20m%5E2%20C%5E%7B-2%7D)
is the Coulomb's constant,
![q_1=q_2=29~nC=29\cdot 10^{-9}~C](https://tex.z-dn.net/?f=q_1%3Dq_2%3D29~nC%3D29%5Ccdot%2010%5E%7B-9%7D~C)
is the intensity of the two charges, and
![r=5.5~cm=0.055~m](https://tex.z-dn.net/?f=r%3D5.5~cm%3D0.055~m)
is the distance between them.
Substituting these numbers into the equation, we get
![F=2.5~10^{-3}~N](https://tex.z-dn.net/?f=F%3D2.5~10%5E%7B-3%7D~N)
The force is repulsive, because the charges have same sign and so they repel each other.
Answer:
Check the explanation
Explanation:
The potential energy <em><u>(is the energy by virtue of a particular object's location relative to that of other objects. This term is most of the time linked or associated with restoring forces such as a spring or the force of gravity.) </u></em>seems to be U = mgy
. In fact, the mgy term has disappeared from the development.
Kindly check the attached image below for the step by step explanation to the question above.