Question

A uniformly charged rod having charge Q and length l is placed at a distance d from another point charge. Find the force of attraction between them.​

Answer 1

Given:

Uniformly charged rod Q is placed at a distance d from a point charge q. Length of rod is l .

To find:

Net Electrostatic Force of attraction

Calculation:

Let us consider an elemental portion on charged rod having charge dQ and length dx. It is located at a distance "x" from Point charge q.

Force due to elemental part of rod be dF

$\therefore \: dF = \dfrac{k(q)(dQ)}{ {x}^{2} }$

$= > dF = \dfrac{k(q) \bigg \{ \dfrac{Q}{l} \times dx \bigg \}}{ {x}^{2} }$

$= > dF = \dfrac{kQq}{l} \bigg \{ \dfrac{dx}{ {x}^{2} } \bigg \}$

Integrating on both sides , we get :

$\displaystyle= > \int dF = \dfrac{kQq}{l} \int \dfrac{dx}{ {x}^{2} }$

Putting the Limits :

$\displaystyle= > \int_{0}^{F} dF = \dfrac{kQq}{l} \int_{d}^{d+l} \dfrac{dx}{ {x}^{2} }$

$= > F = \dfrac{kQq}{l} \bigg \{ \dfrac{ - 1}{x} \bigg \}_{d}^{d+l}$

$= > F = \dfrac{kQq}{l} \bigg \{ \dfrac{1}{d} - \dfrac{1}{d + l} \bigg \}$

$= > F = \dfrac{kQq}{l} \bigg \{ \dfrac{d + l - d}{d(d + l)} \bigg \}$

$= > F = \dfrac{kQq}{ \cancel l} \bigg \{ \dfrac{ \cancel l }{d(d + l)} \bigg \}$

$= > F = \dfrac{kQq}{d(d + l) }$

Putting the value of Coulomb's Constant in place of k , we get :

$= > F = \dfrac{1}{4\pi\epsilon_{0}} \dfrac{Qq}{d(d + l) }$

So final answer is :

$\boxed{ \blue{ \bold{ \huge{F = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{\dfrac{Qq}{d(d + l) } \bigg \} }}}}$