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Answer:
On that line segment between the two charges, at approximately away from the smaller charge (the one with a magnitude of
,) and approximately
from the larger charge (the one with a magnitude of
.)
Explanation:
Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.
Let denote the Coulomb constant, and let
denote the size of a point charge. At a distance of
away from the charge, the electric field due to this point charge will be:
.
At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.
Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.
When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.
On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.
Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.
Let and
. Assume that the electric field is zero at
meters to the right of the
point charge. That would be
meters to the left of the
point charge. (Since this point should be between the two point charges,
.)
The electric field due to would have a magnitude of:
.
The electric field due to would have a magnitude of:
.
Note that at all point in this section, the two electric fields and
will be acting in opposite directions. At the point where the two electric fields balance each other precisely,
. That's where the actual electric field is zero.
means that
.
Simplify this expression and solve for :
.
.
Either or
will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges,
. Therefore,
isn't a valid value for
in this context.
As a result, the electric field is zero at the point approximately away the
charge, and approximately
away from the
charge.
Answer:
The position of stable equilibrium is -a
And the period of small oscillations must be: c/(ma^3)
Explanation:
Since the potential is:
We first look for a position of stable equilibrium. This posiiton must satisfy two considtions, that the first derivative of the potential must vanish at this point and the second derivative must be positive.
Which vanish for
x = a ; x =-a
The second derivative of V(x) is:
And:
Therefore:
a)
The position of stable equilibrium is -a
And the period of small oscillations must be:
(c/(ma^3))^1/2
b)
Let's find the maximum amplitude if the particle starts at this point with velocity v
If this is the case, the total energy will be:
(mv^2)/2
And the maximum amplitude will be
x = a^3/c mv^2 = (m v^2 a^3)/ c