Three tiny charged metal balls are arranged on a straight line. The middle ball is positively charged and the two outside balls
are negatively charged. The two outside balls are separated by 20 cm and the middle ball is exactly halfway in between. (HINT: Draw a picture; in your picture, the distance between the two outermost balls should be 20 cm.) The absolute value of the charge on each ball is the same, 1.12 μCoulombs (the meaning of μ, which is read as "micro", is 10-6). Give your answers in newtons.
(a) What is the magnitude of the attractive force on either outside ball due ONLY to the positively-charged middle ball? N
(b) What is the magnitude of the repulsive force on either outside ball due ONLY to the other outside ball? N
(c) What is the magnitude of the net force on either outside ball? N
(a) What is the magnitude of the attractive force on either outside ball due ONLY to the positively-charged middle ball? N
(b) What is the magnitude of the repulsive force on either outside ball due ONLY to the other outside ball? N
(c) What is the magnitude of the net force on either outside ball? N
1 answer:
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a) E = 0
b)
Explanation:
a)
We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:
where
E is the electric field
q is the charge contained by the Gaussian surface
is the vacuum permittivity
Here we want to find the electric field at a distance of
r = 12 cm = 0.12 m
Here we are between the inner radius and the outer radius of the shell:
However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.
This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:
q = 0
Therefore, the magnitude of the electric field is also zero:
E = 0
b)
Here we want to find the magnitude of the electric field at a distance of
r = 20 cm = 0.20 m
from the centre of the shell.
Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.
Therefore, it is given by:
where in this problem:
is the charge on the shell
is the distance from the centre of the shell
Substituting, we find: