A line of charge starts at x = x0, where x0 is positive, and extends along the x-axis to positive infinity. If the linear charge density is given by λ = λ0 x0/x, where λ0 is a positive constant, find the electric field at the origin. (ˆı denotes the unit vector in the positive x direction.)

E = 0.5*k*λ_o/x_o

Step-by-step explanation:

Given:

- A line of charge with a variable charge density:

                                     λ = λ_o*x_o/x

- The line of charge starts from x = x_o to +infinity.

Find:

Find the electric field at the origin.

Solution:

- We will first develop an expression of a differential section of the line that causes an electric field at the origin as follows:

                                   dE = k*λ.dx / x^2

- Remember that λ is a function of x as given in the expression. So will plug the expression of charge density into the differential form:

                                  dE = (k*λ_o*x_o/x^3). dx

- Now, we will sum up all the differential elements of the line of charge. We will integrate the expression from x = x_o to x =+inf.

                                  E = k*λ_o*x_o integral ( 1 / x^3 ) .dx

                                  E = -0.5*k*λ_o*x_o ( 1 / x^2)

Evaluate the limits:

                                  E = -0.5*k*λ_o*x_o ( 0 - 1/x_o^2)

                                  E = 0.5*k*λ_o/x_o

answer: 21$

step-by-step explanation:

$42 / $14 = 3   (ratio 3: 1)

$7 x 3 = $21

the answer is b.

step-by-step explanation:

by the trigonometric ratio tangent, tangent equals opposite divided by adjacent.

therefore, tan(50) equals ac/3

multiply each side by 3 to get: ac=3tan(50)

the answer is b.