Answer:
Electric field, E = 936.19 N/C
Explanation:
It is given that,
Charge 1, 
Charge 2, 
Distance between them, d = 3 mm = 0.003 m
Torque, 
Angle between electric field and line connecting the charge, 
We need to find the torque exerted on the dipole. The torque experienced by the dipole in the electric field is given by :
p is the dipole moment, 
E = 936.19 N/C
So, the magnitude of electric field on the dipole is 936.19 N/C. Hence, this is the required solution.
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Answer:
2000 J
Explanation
Work equation is expressed as:
Where:
F: Applied force
d: traveled distance
α: Angle between the direction of the force and the direction of the movement. (in this case, both of the direction are the same, so the angle is 0°)
By substituting:
Have a nice day!
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Answer:
this is violation of II law of thermodynamics
Explanation:
As per 2nd law of thermodynamics we know that heat always flows from high temperature to low temperature
If heat flows from low temperature to high temperature then it is only possible when some external work is done on the system
So here it says that initially two blocks are at 10 degree C and 20 degree C
Now the cool object will become cooler and hot object becomes hotter
which shows that heat is flowing from low temperature to high temperature
So this is violation of II law of thermodynamics
Answer:
The electric field at x = 3L is 166.67 N/C
Solution:
As per the question:
The uniform line charge density on the x-axis for x, 0< x< L is 
Total charge, Q = 7 nC = 
At x = 2L,
Electric field, 
Coulomb constant, K = 
Now, we know that:
Also the line charge density:
Thus
Q = 
Now, for small element:
Integrating both the sides from x = L to x = 2L
![\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]](https://tex.z-dn.net/?f=%5Cvec%7BE_%7B2L%7D%7D%20%3D%20K%5Clambda%5B%5Cfrac%7B-%201%7D%7Bx%7D%5D_%7BL%7D%5E%7B2L%7D%5D%20%3D%20K%5Cfrac%7BQ%7D%7BL%7D%5Bfrac%7B1%7D%7B2L%7D%5D)
![\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}](https://tex.z-dn.net/?f=%5Cvec%7BE_%7B2L%7D%7D%20%3D%20%289%5Ctimes%2010%5E%7B9%7D%29%5Cfrac%7B7%5Ctimes%2010%5E%7B-%209%7D%7D%7BL%7D%5Bfrac%7B1%7D%7B2L%7D%5D%20%3D%20%5Cfrac%7B63%7D%7BL%5E%7B2%7D%7D)
Similarly,
For the field in between the range 2L< x < 3L:
![\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20K%5Clambda%5B%5Cfrac%7B-%201%7D%7Bx%7D%5D_%7B2L%7D%5E%7B3L%7D%5D%20%3D%20K%5Cfrac%7BQ%7D%7BL%7D%5Bfrac%7B1%7D%7B6L%7D%5D)
![\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%289%5Ctimes%2010%5E%7B9%7D%29%5Cfrac%7B7%5Ctimes%2010%5E%7B-%209%7D%7D%7BL%7D%5Bfrac%7B1%7D%7B6L%7D%5D%20%3D%20%5Cfrac%7B63%7D%7B6L%5E%7B2%7D%7D)
Now,
If at x = 2L,
Then at x = 3L:
