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:
the cost of operating the light bulb is 72 cents.
Explanation:
Given;
cost of electricity, C = 6 cents / kW.h
power of light bulb, P = 100 W
time of light power consumption, t = 4 hours per day for 30 days
total time = 4 hours x 30 = 120 hours
Power consumed by the light bulb is calculated as;
P = 100 x 120 = 12000 w.h = 12 kW.h
Cost of power consumption = 6 cents/kWh x 12 kWh
= 72 cents
Therefore, the cost of operating the light bulb is 72 cents.
Answer:
-730KJ
Explanation:
According to the first law of thermodynamics;
Let the total energy of the system be ∆E
Let heat be q and let work the w
Since the energy decreases ∆E is negative
Since work is done on the system w= positive
So;
- 250 = 480 + q
q = -250 - 480
q=-730KJ
The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.
Answer:Given:
Initial speed of fugitive, v0 = 0 m/s
Final speed, vf = 6.1 m/s
acceleration, a = 1.4 m/s^2
Speed of train, v = 5.0 m/s
Solution:
t = (vf-v0)/a
t = (6.1-0)/1.4
t =4.36 s
Distance traveled by train, x_T =v*t
x_T =5*4.36 = 21.8 m
Distance travelled by fugitive, x_f = v0*t+1/2at^2
x_f = 0*4.36+1/2*1.4*4.36^2
x_f =13.31 m
5*t = v(t-4.36)+x_f
5*t=6.1*(t-4.36)+13.31
solve for t, we get
t = 12.08 s
The fugitive takes 12.08 s to catch up to the empty box car.
Distance traveled to reach the box car is
X_T = v*t
X_T = 5*12.08 s
X_T = 60.4 m
Explanation:
