In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point
1 answer:
The difference in electric potential energy between the two points is
where q is the magnitude of the charge and
is the electric potential difference.
But for energy conservation, the difference in electric potential energy
between the two points is equal to the work done to move the charge between A and B:
so we have
and by substituting the numbers of the problem, we find the value of:
where q is the magnitude of the charge and
But for energy conservation, the difference in electric potential energy
so we have
and by substituting the numbers of the problem, we find the value of
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When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
(2)
where
is the refractive index of the second medium and 
is the refractive index of the first medium.
We can find the ratio
by using Snell's law:
(1)
where
is the angle of incidence
is the angle of refraction
By using the data of the problem and re-arranging (1), we find
and if we use eq.(2) we can now find the value of the critical angle:
where
We can find the ratio
where
By using the data of the problem and re-arranging (1), we find
and if we use eq.(2) we can now find the value of the critical angle:
In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
where t is the time,
is the maximum charge on the capacitor at voltage V, and 
is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.
1) Using
and 
, the time constant of the circuit is:
2) To find the charge on the capacitor at time
, we must find before the maximum charge on the capacitor, which is
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3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be
. We can see this also from the previous formule, by using 
:
where t is the time,
Using this law, we can answer all the three questions of the problem.
1) Using
2) To find the charge on the capacitor at time
And then, the charge at time
3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be