A small charge q is placed near a large spherical charge Q. The force experienced by both charges is F. The electric eld created
1 answer:
The electric field created by Q at the position of q is .
The given parameters:
- <em>Magnitude of charge, = q</em>
- <em>Spherical charge, = Q</em>
- <em>Force experienced by both charges, = F</em>
The electric field created by Q at the position of q is calculated as follows;
where;
- <em>E is the magnitude of electric field strength </em>
- <em>F is the force experienced by both charges</em>
- <em>Q is the charge</em>
Thus, the electric field created by Q at the position of q is .
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Answer:
Electric field at radius r inside the solid sphere is
Electric field at radius r between inner radius and outer radius of the shell is
E=0 N/C
Explanation:Given
The radius of the solid sphere is
The charge on the solid sphere is
The inner radius of the shell is
The outer radius of the shell is
The total charge on the shell is
PART(A)
The magnitude of electric field at radius r where \\The volumetric charge density of the solid sphere will be
The charge enclosed by the radius r inside the solid sphere is
rho=
According to gauss law
PART(B)
The electric field at radius r where
The shell is conducting so due to induction of charge
The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell
So,
The charge on the inner surface of the shell is
Due to conservation of the charge on the shell
The charge accumulated on the outer surface of the shell is
The charge enclosed by the radius r where
According to gauss law
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Answer:
Explanation:
From, the given information: we are not given any value for the mass, the proportionality constant and the distance
Assuming that:
the mass = 5 kg and the proportionality constant = 50 kg
the distance of the mass above the ground x(t) = 1000 m
Let's recall that:
Similarly, The equation of mption:
replacing our assumed values:
where
So, when the object hits the ground when x(t) = 1000
Then from above derived equation:
By diregarding
1000 + 0.981 = 0.981 t
1000.981 = 0.981 t
t = 1000.981/0.981
t = 1020.36 sec