Question

Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positivex=axis at x = a+r, a distance r to the right end of Q.(a) Calculate the x- and y-components of the electric fieldporduced by the charge distribution Q at points on the positivex-axis where x>a.(b) Calculate the force (magnitude and direction) that thecharge distribution Q exerts on q.(c) Show that if r>>a, the magnitude of the force inpart (b) is approximately Qq/(4pi epsilon0 r^2). Explain why thisresult is obtained.

Answer 1

a. b- x= y

dx = -dy

b. F =
`(-kQqi)/(r (a+r))`

c. F =
`(-kQqi)/(r^(2) )`

Step-by-step explanation:

a. x components:

`dE = (kdq)/((a+r-x^(2)) ) \\`

=
`(kQdx)/((a(a+r-x)^2)`

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F =
`(-kQqi)/(r (a+r))`

c. Given that r>>a, the expression becomes:

F =
`(-kQqi)/(r^(2) )`

Step-by-step explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.