Question

determine th e electric intensity due to a long wire of uniform intensity charge density

Answer 1

Question:

• Determine the electric field intensity at any point on the axis of a uniformly charged long wire having uniform charge density.

Answer:

$\large\bold\red{E = \frac{ \lambda}{4\pi \epsilon_ {0} } \: \frac{l}{a(l + a)} }$

Explanation:

Let the length of wire be l.

And,

It's linear charge density be $\bold{\lambda}$.

$\bold\purple{Note:-}Refer\: to\; the\: attachment\: for\: figure$

Now,

Consider an element ‘dx’ at a distance x from the point P, where we seek to find the electric field.

Therefore,

The elemental charge ,

$\boxed{ \bold{dq = \lambda \: dx}}$

Therefore,

We have,

$= > dE = k \: \frac{ \lambda \: dx}{ {x}^{2} }$

Integrating both the sides,

We get,

$= > \displaystyle\int \: dE = k \lambda \: \displaystyle \int \limits_ {a}^{a + l} \frac{1}{ {x}^{2} } dx \\\\ \\ = > E = \displaystyle{k \lambda| - \frac{1}{x} | \displaystyle{_{a}^{a + l} }} \\ \\ \\ = > E = k \lambda(- \frac{1}{a + l } + \frac{1}{a} ) \\ \\ \\ = > E = \frac{ \lambda}{4\pi \epsilon_ {0} } \: ( \frac{1}{a} - \frac{1}{l + a} ) \\\\ \\ = > \large \boxed{ \bold{ E = \frac{ \lambda}{4\pi \epsilon_ {0} } \: \frac{l}{a(l + a)} }}$