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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:
Answer:
Work done on the box = 32 J
Explanation:
Work done = Force x Displacement.
Force applied on the box = 16 N
Displacement of box = 2 meter.
Work done on the box = Force applied on the box x Displacement of box
= 16 x 2 = 32 Nm = 32 J
Work done on the box = 32 J
I believe it's the the third one. :)
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Answer:
166.67 N
Explanation:
Applying Pascal's principle,
Presure in the smaller piston(P') = Pressure in the bigger piston(P)
But,
Pressure = Force/Area
Pressure in the smaller piston(P') = Force applied to the smaller piston(F')/Area of the smaller piston(A')
Pressure in the bigger piston(P) = Force applied to the bigger piston(F)/Area of the bigger piston(A)
F'/A' = F/A.................. Equation 1
Make F the subject of the equation
F = F'A/A'.............. Equation 2
From the question,
Given: F' = 20 N, A = 120 cm², A' = 10 dm² = (10×100) = 1000 cm²
Substitute these values into equation 2
F = 20(1000)/120
F = 166.67 N
