Two charged spheres are 20 cm apart and exert an attractive force of 8 x 10-9 n on each other. What will the force of attraction
1 answer:
Answer:
Explanation:
The inital electrostatic force between the two spheres is given by:
where
is the initial force
k is the Coulomb's constant
q1 and q2 are the charges on the two spheres
r is the distance between the two spheres
The problem tells us that the two spheres are moved from a distance of r=20 cm to a distance of r'=10 cm. So we have
Therefore, the new electrostatic force will be
So the force has increased by a factor 4. By using , we find
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Answer:
a) fem = - 2.1514 10⁻⁴ V, b) I = - 64.0 10⁻³ A, c) P = 1.38 10⁻⁶ W
Explanation:
This exercise is about Faraday's law
fem =
where the magnetic flux is
Ф = B x A
the bold are vectors
A = π r²
we assume that the angle between the magnetic field and the normal to the area is zero
fem = - B π 2r dr/dt = - 2π B r v
linear and angular velocity are related
v = w r
w = 2π f
v = 2π f r
we substitute
fem = - 2π B r (2π f r)
fem = -4π² B f r²
For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T
we reduce the magnitudes to the SI system
f = 2 rev / s (2π rad / 1 rev) = 4π Hz
we calculate
fem = - 4π² 428 10⁻⁶ 4π 0.10²
fem = - 16π³ 428 10⁻⁶ 0.010
fem = - 2.1514 10⁻⁴ V
for the current let's use Ohm's law
V = I R
I = V / R
I = -2.1514 10⁻⁴ / 0.00336
I = - 64.0 10⁻³ A
Electric power is
P = V I
P = 2.1514 10⁻⁴ 64.0 10⁻³
P = 1.38 10⁻⁶ W
Answer:
B = 2.5 10⁸ Pa
Explanation:
The volume modulus is defined by
B =
The negative fate is for the module to be positive since the volume change is negative
It is not necessary to reduce the volumes to the SI system, since they are both in the same units
B = =
B = 2.5 10⁸ Pa