contestada

A charge Q= 2 C is distributed uniformly through out a bar of length L=2.5 m. The bar is placed horizontally in free space. A second charge q = 10−9C is placed along the line of the bar a distance d= 2m away in space, measured from the right end of the bar. What force is exerted on charge q by the charged bar?

Respuesta :

Answer:

F = 2 N

Explanation:

Let the rod is made up of large number of point charges

so here force due to one small part which is at distance "x" from the end of the rod on the given charge is

[tex]dF = \frac{kdq q}{(d + x)^2}[/tex]

here we know that

[tex]dq = \frac{Q}{L} dx[/tex]

so we have

[tex]F = \int \frac{k Q q dx}{L(d + x)^2}[/tex]

[tex]F = \int_0^L \frac{kQq dx}{L(d + x)^2}[/tex]

[tex]F = \frac{kQq}{L} ( \frac{1}{d} - \frac{1}{d + L})[/tex]

now plug in all data

[tex]F = \frac{(9\times 10^9)(2C)(10^{-9})}{2.5} (\frac{1}{2} - \frac{1}{(2 + 2.5)})[/tex]

[tex]F = 2N[/tex]