Physics, asked by patrickgerald16, 6 months ago

this is a question of physics class 12th chapter 1 electric charges and fields. 1) find electric field due to long wire having uniform linear charge density by using Gaus's law ​

Answers

Answered by jigglepiggle12091209
1

Explanation:

so from the formula when charge density gets double and the distance between point and line of charge is half, the electric field become 4 times.

Answered by aviraltiwari12516
1

ANSWER

Take a long thin wire of uniform linear charge density λ.

Consider a point A at a perpendicular distance l from the mid-point O of the wire, as shown in the following figure.

Let E be the electric field at point A due to the wire, XY.

Consider a small length element dx on the wire section with OZ=x

Let q be the charge on this piece.

∴q=λdx

Electric field due to the piece,

dE=

4π∈

0

1

(AZ)

2

λdx

However, AZ=

(l

2

+x

2

)

∴dE=

4π∈

0

(l

2

+x

2

)

λdx

The electric field is resolved into two rectangular components. dEcosθ is the perpendicular component and dEsinθ is the parallel component.

When the whole wire is considered, the component dEsinθ is cancelled.

Only the perpendicular component dEcosθ affects point A.

Hence, effective electric field at point A due to the element dx is dE

1

.

∴dE

1

=

4π∈

0

(x

2

+l

2

)

λdxcosθ

... (1)

In △AZO,

tanθ=

l

x

x=ltanθ ... (2)

On differentiating equation (2), we obtain

dx=lsec

2

(θ)dθ

dE

1

=

4πε

o

l

λcos(θ)dθ

E

1

=∫

−π/2

π/2

4πε

o

l

λcos(θ)dθ

Solving, E

1

=

2πε

o

l

λ

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