Answer:
Explanation:
A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Let given is,
The diameter of a parallel plate capacitor is 6 cm or 0.06 m
Separation between plates, d = 0.046 mm
The potential difference across the capacitor is increasing at 500,000 V/s
We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :
, r is radius
Let I is the displacement current. It is given by :
Here, 
is rate of increasing potential difference
So
So, the value of displacement current is 
.
It will be unaffected by the magnet because it has no magnetic field. If you were to maybe have electricity going through it is the only way it would have anything to do with the magnet.
<span />
<u>Halfway</u><u> between the like poles of two magnets, because the field lines bend away and do not enter this area.</u>
How does a magnetic field diagram show where the field is strongest?
- The magnetic field lines do not ever cross.
- The lines include arrowheads to indicate the direction of the force exerted by a magnetic north pole.
- The closer the lines are to the poles, the stronger the magnetic field (thus the magnetic field from a bar magnet is highest closest to the poles).
Where is magnetic field the strongest and weakest on a magnet?
- The bar magnet's magnetic field is strongest at its core and weakest between its two poles.
- The magnetic field lines are densest immediately outside the bar magnet and least dense in the core.
Which two locations on the magnet would have the greatest attractive forces?
- Inside the magnet itself, the field lines run from the south pole to the north pole.
- The magnetic field is strongest in areas of greatest density of magnetic field lines, or areas of the greatest magnetic flux density.
Learn more about magnetic field
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Electrostatic forces between charges depend on the product of
the sizes of the charges, and the distance between them.
We should also mention the item about whether the charges are
both the same sign or opposite signs. That determines whether
the forces will pull them together or push them apart, which is a
pretty significant item.
<span>The formulas are,
v1d1² = v2d2² ........ (1)
h = (v2²-v1²)/2g ...... (2)
Given that,
v1 = 1.71 m/s
we assume that the stream has decreased by a factor
d2 =0.805d1
then,
v1d1² = v2 (0.805d1)²
cancelled both side d1² then we get,
v1 = v2 (0.805)²
v1 = v2 (0.648025)
Sub v1 = 1.71,
1.71 = v2 (0.648025)
v2 = 1.71/0.648025
v2 = 2.638787083831642
v2 = 2.64 m/s
The vertical distance formula,
h = (v2²-v1²)/2g
We know that value of gravity constant is 9.8 m/s²
h = {(2.64)² - (1.71)²)/2(9.8)
h = {(6.9696) - (2.9241)}/19.6
h = (4.0455)/19.6
h = 0.2064030612244898
h = 0.21 cm
Therefore, the vertical distance h = 0.21 cm.</span>
