Question

A uniformly charged rod of length L = 1.3 m lies along the x-axis with its right end at the origin. The rod has a total charge of Q = 2.2 μC. A point P is located on the x-axis a distance a = 1.8 m to the right of the origin.

Answer 1

Answer:

E = 3544.44 N/C

Explanation:

Given:

- charge Q = 2.2 *10^-6 C

- Length L = 1.3 m

Find:

The Electric Field strength E @ a  = 1.8 m

Solution:

- The differential electric field dE due to infinitesimal charge dq can be considered as a point charge at a distance of r is given by:

                                  dE = k*dq / r^2

- The charge Q is spread over entire length L, hence:

                                  dq = (Q / L ) * dx

-The resulting dE:

                                 dE = (k*Q/L)*(dx / r^2)

- point P lies on the x- axis with distance (x+a) from differential charge from:

                                 dE = (k*Q/L)*(dx / (x+a)^2)  

- Integrate dE over length 0 to L

                                 E = (-k*Q/L)*( 1 / (x+a) )

                                 E =  (-k*Q/L)* (1 / a - 1 / (L+a))

                                 E =  (-k*Q/L)* (L / a(L+a))

                                 E = (k*Q / a(L+a))

- Evaluate E @ a = 1.8 m

                                 E =(8.99*10^9 * 2.2*10^-6 / 1.8*(1.3+1.8))

                                 E = 3544.44 N/C