00:01
For this problem on the topic of electrostatics, we have a charge q1, which is plus 2q at the origin, and a charge q2, which is minus q on the x -axis at x is equal to a.
00:11
And we want to find the expressions for the total electric field on the x -axis in each of the regions, x less than 0, x between 0 and a and x greater than a, and then determine all the points on the x -axis where the electric field is 0.
00:23
We then want to make a plot of e versus x for all points on the x axis and qualitatively discuss what happens between minus infinity and infinity.
00:36
Now, firstly, for part a, we want the x component of the field x, and this is equal to the x component of the field due to 1 plus the x component of the field due to charge 2.
00:52
And this is equal to minus k -k -1 over x squared plus k -2 over a plus x all squared.
01:09
And we can write this as k -into minus q -1 over x squared plus the magnitude of charge 2, q2 over the distance to the point a plus x all squared.
01:29
And so we can then write this as k into minus 2q over x squared plus q over a plus x all squared.
01:47
And so we have an expression for the electric field in the first region.
01:55
In the next region, we have the electric field x is equal to, again, e1x plus e2x, which in this case is k times q1 over x squared plus k times q2 over a minus x or squared.
02:22
This is equal to k into the magnitude of q1 divided by x squared plus the magnitude of q2 divided by a minus x all squared.
02:38
Putting in the value for the charges, this is k into 2k over x squared plus q divided by a minus x all squared.
02:57
And finally, for the third region, for part c, we again have the net field, x, to be the contribution due to charge 1 plus the contribution due to charge 2...