Question

apply gauss's law to derive the expression for electric field intensity due to an infinitely long straight uniformly charged wire.what is the direction of the field intensity if it is positively charged.​

Answer 1

Answer:

E= ( sigma / 2€° ) r^

Explanation:

1. Direction of field intensity is radially outward if charge is positive.....

Answer 2

Answer :-

$\to \boxed{ \sf{ E = \frac{ \lambda}{ 2 \pi \: r \:\epsilon_{o} \:} }} \:$

To Find :-

→ Electric field intensity and direction of it .

Explanation :-

$\overline{ \underline{ \large{ \bf{ \red{Gauss }\: \green{Law } }\: } }} : - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{it \: states \: that \: the \: electric \: } \\ \sf{flux \: through \: any \: closed \: surface \: is \: equal \: to \: } \\ \sf{ \frac{1}{ \epsilon_{o}} \: of \: \: distributed \: charge \: .} \\ \\ \pink{\implies} \boxed{ \sf{ \phi_{e} = \frac{q}{\epsilon_{o}}}}$

And, this closed surface is known as Gaussian surface .

This intensity is equal to closed area of electric field and ds(small area ) ( with dot product)

We know that ;

$\to \boxed{ \sf{\vec{a}. \vec{b} = a \: b \cos \theta \: }} \\ \\ \implies \sf{ \phi_{e} = \oint \: E.ds} \\ \\ \implies \: \sf{\phi_{e} = \oint \: E.ds \: \cos \theta} \\ \\ \because \theta = 0 \: so \: \cos \theta = 1 \\ \\ \to \sf{\phi_{e} = \: E \oint \: ds \: } \\ \: \\ \to \boxed{ \sf{ \phi_{e} = \: E \times 2 \pi \: r \: l\: }}\\ \:$

compare these intensity ,

$\to \sf{ E \times 2 \pi \: r \: l= \frac{q}{\epsilon_{o} } } \\ \\ \to \sf{ E = \frac{1}{ 2 \pi \: r \: l} \times \frac{q}{\epsilon_{o}} } \\ \\ \because \boxed{ \sf{ \lambda = \frac{ q}{l} \to \: q = \lambda \: l}} \\ \\ \to \: \sf{ E = \frac{1}{ 2 \pi \: r \: l} \times \frac{ \lambda \: l}{\epsilon_{o}} } \\ \\ \to \boxed{\sf{ E = \frac{1}{ 2 \pi \: r \: } \times \frac{ \lambda}{\epsilon_{o}} } \: } \\ \\ \to \boxed{ \sf{ E = \frac{ \lambda}{ 2 \pi \: r \:\epsilon_{o} \:} }} \\ \\ \to \bf{ \lambda \: is \: linear \: charge \: density \: }$