There is a force between two charges; if the distance between the two charges is doubled, by what factor does the force between
2 answers:
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Answer:
P = 7.28 N.s
Explanation:
given,
initial momentum of cue ball in x- direction,P₁ = 9 N.s
momentum of nine ball in x- direction, P₂ = 2 N.s
momentum in perpendicular direction i.e. y - direction,P'₂ = 2 N.s
momentum of the cue after collision = ?
using conservation of momentum
in x- direction
P₁ + p = x + P₂
p is the initial momentum of the nine balls which is equal to zero.
9 + 0 = x + 2
x = 7 N.s
momentum in x-direction.
equating along y-direction
P'₁ + p = y + P'₂
0 + 0 = y + 2
y = -2 N.s
the momentum of the cue ball after collision is equal to resultant of the momentum .
P = 7.28 N.s
the momentum of the cue ball after collision is equal to P = 7.28 N.s
When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling the capacitance of the capacitor in air, the charge Q, the capacitance
and the voltage (
) are related by
(1)
when the source is disconnected the charge Q remains on the capacitor.
When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):
this is the new capacitance. Since the charge Q on the plates remains the same, by using eq. (1) we can find the new voltage across the capacitor:
And since , substituting into the previous equation, we find: