Answer:
(a) 6650246.305 N/C
(b) 24150268.34 N/C
(c) 6408227.848 N/C
(d) 665024.6305 N/C
Explanation:
Given:
Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]
Total charge of the ring (Q) = 75.0 μC = 
[1 μC = 10⁻⁶ C]
Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:
Plug in the given values for each point and solve.
(a)
Given:
, 
Electric field is given as:
(b)
Given:
, 
Electric field is given as:
(c)
Given:
, 
Electric field is given as:
(d)
Given:
, 
Electric field is given as:
In order to delete a file, it is required to first select the file which you want to delete and then select 'delete' from the file menu. It will then ask you to confirm the deletion which will be required in order to complete the file deletion process.
It isn't required to click on the X at the top right of the screen or to open the file you want to delete as these steps aren't relevant to the deletion process.
Hence, the statements which describe the steps you need to take in order to delete a file are as follows:
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Answer:
V = 192 kV
Explanation:
Given that,
Charge, 
Distance, r = 0.3 m
We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :
So, the required electric potential is 192 kV.
Answer:
57,42 KJ
Explanation:
By a isobaric proces, the expresion for the works in the jpg adjunt. Then:
W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)
By the ideal gases law: PV=RTn
Then, in (1): (remember Pa = Pb)
W = R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)
Since we have 1 Kg air: How much is this in moles?
From bibliography: 28.96 g/mol
Then, in 1 Kg (1000 g) there are:
n = 34,53 mol
Finally, in (2):
W = (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ
