Given a line of charge in the x-axis which exists the from a to b with a charge distribution of lambda(x)= Cx(X_o-x)^2, calculate the total electric field at x_o. (X_o, C, a and b are positive; a<b<x_o)
To calculate the total electric field at x_o due to the line of charge with a given charge distribution, you need to consider the contribution from infinitesimally small charge elements along the line.
The electric field created by an infinitesimally small charge element is given by Coulomb's law:
dE = k * dq / r^2 * r̂,
where dE is the infinitesimal electric field produced by the charge element, k is the electrostatic constant (k = 9 × 10^9 N m^2/C^2), dq is the charge element, r is the distance between the charge element and the point where the electric field is being calculated, and r̂ is the unit vector pointing from the charge element to the point of interest.
To find the electric field at x_o, we need to integrate the contributions from all the charge elements along the line. The total electric field will be the vector sum of all these infinitesimal electric fields.
Let's proceed step by step:
1. Express the charge element dq in terms of the charge distribution lambda(x).
dq = lambda(x) * dx.
Here, dx is the infinitesimal length element along the x-axis.
2. Express the infinitesimal electric field dE in terms of dq and r.
dE = k * dq / r^2 * r̂.
3. Express r in terms of x and x_o.
r = x_o - x.
4. Substitute the expressions for dq and r into the equation for dE.
dE = k * lambda(x) * dx / (x_o - x)^2 * r̂.
5. Integrate the electric field expression over the length of the line charge from a to b to obtain the total electric field at x_o.
E(x_o) = ∫[a to b] k * lambda(x) * dx / (x_o - x)^2 * r̂.
Now, plug in the given charge distribution and evaluate the integral to find the total electric field at x_o.