Answer:
The magnitude of the electric field at a point equidistant from the lines is
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle
We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field
Put the value into the formula
Hence, The magnitude of the electric field at a point equidistant from the lines is
13200N
Explanation:
Given parameters:
Mass = 1100kg
Velocity = 24m/s
time = 2s
unknown:
Braking force = ?
Solution:
The braking force is the force needed to stop the car from moving.
Force = ma =
m is the mass of the car
v is the velocity
t is the time taken
Force = = 13200N
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The elastic potential energy of the spring is 6.8 J
Explanation:
The elastic potential energy of a compressed/stretched spring is given by the equation:
where
k is the spring constant
x is the elongation of the spring
The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):
By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,
Therefore when the spring has a elongation of , its potential energy is
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a) Electric field inside the paint layer: zero
b) Electric field just outside the paint layer:
c) Electric field 8.00 cm outside the paint layer:
Explanation:
a)
We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:
where
E is the electric field
dS is the element of surface
q is the charge within the gaussian surface
is the vacuum permittivity
Here we want to find the electric field just inside the paint layer, so we take a sphere of radius as Gaussian surface, where
R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)
By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore
So we have
which means that the electric field inside the paint layer is zero.
b)
Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius
The area of the surface is
And since the electric field is perpendicular to the surface at any point, Gauss Law becomes
The charge included within the sphere in this case is the charge on the paint layer, therefore
So, the electric field is:
where the negative sign means the direction of the field is inward, since the charge is negative.
c)
Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.
Therefore, we have to take a Gaussian sphere of radius:
Gauss theorem this time becomes
And the charge included within the sphere is again the charge on the paint layer,
Therefore, the electric field is
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