What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 47.0 nC placed between q1
and q2 at x3 = -1.075 m ? Your answer may be positive or negative, depending on the direction of the force.
1 answer:
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Answer:
For the point inside the cylinder:
For the point outside the cylinder:
where x0 is the position of the point on the x-axis and σ is the surface charge density.
Explanation:
Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.
<u>x = -x0:</u>
The cylinder is consist of the sum of the rings with the same radius.
First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.
We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.
The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.
Now, the electric field created by the small portion is 'dE'.
The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.
We have to integrate it over the ring, which is an angular integration.
This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:
The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.
<u>x = +x0: </u>
We will only change the boundaries of the last integration.