00:01
Okay, so we have the e is equal to keq over x times x plus 2a.
00:07
So this is the electric field at a point a distant x from one end of a line charge of lens 2a.
00:17
So therefore, when the de can be equal to ke times dq over x times x plus 2a, then i have df, which is the derivative of the force, is equal to charge q times dc, which is equal to k -e -k -tqued -d -k over x times x plus 2a.
00:37
And we know charge q is equal to charge per union length times the length of raw, which is 2a.
00:44
And then it will give us d -q is equal to lambda times dx.
00:47
And dx is the infinitesimal segment, okay? so therefore df is equal to k -e -k times lambda times d x over x times x plus 2a.
01:01
So therefore we have force is equal to the integral of derivative of the force, which is equal to k -eq londa d x over x times x plus 2a and the ranges from b minus 2a to b since the length of the thin rise 2a, okay? and the distance between the rise b.
01:22
So therefore we have force is equal to k -eq londa times the integral.
01:27
Then we can start solving the question.
01:31
So well we know force, now can be equal to keq, landa, times the anti -derivative of dx over x times x plus 2a, which is negative 1 over 2a times natural log x plus 2a over x, and the range is b minus 2a to b...