Question

The electrical field inside a long cylindrical shell of uniform density is A.Zero B.constant and radial inward C. Directly proportional to the distance from the centre D.Inversely proportional to the square of the distance from the centre​

Answer 1

Given:

Long cylindrical shell of uniform charge density.

To find:

Field intensity inside it?

Calculation:

Let us assume a distance "r" from geometric axis, such that r < R (R is the radius of the cylinder).

• $q_{e}$ refers to enclosed charge within the Gaussian Surface.

Now, applying Gauss' Law:

$\displaystyle \oint \vec{E}. \vec{ds} = \frac{q_{e}}{ \epsilon_{0}}$

$\implies\displaystyle \oint E \times ds \times \cos( {0}^{ \circ} ) = \frac{q_{e} }{ \epsilon_{0}}$

$\implies\displaystyle \oint E \times ds = \frac{q_{e}}{ \epsilon_{0}}$

• Now, as it is a cylindrical shell, the enclosed charge is zero.

$\implies\displaystyle \oint E \times ds = \frac{0}{ \epsilon_{0}}$

$\implies\displaystyle E = 0$

So, field inside the cylindrical shell is zero.

Answer 2

Explanation:

Given:

Long cylindrical shell of uniform charge density.

To find:

Field intensity inside it?

Calculation:

Let us assume a distance "r" from geometric axis, such that r < R (R is the radius of the cylinder).

q_{e}q

e

refers to enclosed charge within the Gaussian Surface.

Now, applying Gauss' Law:

\displaystyle \oint \vec{E}. \vec{ds} = \frac{q_{e}}{ \epsilon_{0}} ∮

E

.

ds

=

ϵ

0

q

e

\implies\displaystyle \oint E \times ds \times \cos( {0}^{ \circ} ) = \frac{q_{e} }{ \epsilon_{0}} ⟹∮E×ds×cos(0

∘

)=

ϵ

0

q

e

\implies\displaystyle \oint E \times ds = \frac{q_{e}}{ \epsilon_{0}} ⟹∮E×ds=

ϵ

0

q

e

Now, as it is a cylindrical shell, the enclosed charge is zero.

\implies\displaystyle \oint E \times ds = \frac{0}{ \epsilon_{0}} ⟹∮E×ds=

ϵ

0

0

\implies\displaystyle E = 0⟹E=0

So, field inside the cylindrical shell is zero.