Question

Electric field intensity due to a uniformly charged solid sphere

Answer 1

To Derive:

Electric field intensity due to a uniformly charged solid sphere.

Solution:

• Let radius of sphere be R :

At a distance r < R:

$\displaystyle\therefore \: \oint E \times ds = \dfrac{q}{ \epsilon_{0}}$

$\displaystyle\implies \: \oint E \times ds = \dfrac{ \bigg \{\dfrac{ q_{0}(\frac{4}{3} \pi {r}^{3} ) }{ \frac{4}{3}\pi {R}^{3} } \bigg \}}{ \epsilon_{0}}$

$\displaystyle\implies \: E \oint \: ds = \dfrac{ \bigg \{\dfrac{ q_{0}(\frac{4}{3} \pi {r}^{3} ) }{ \frac{4}{3}\pi {R}^{3} } \bigg \}}{ \epsilon_{0}}$

$\displaystyle\implies \: E \oint \: ds = \dfrac{ q_{0} \times {r}^{3} }{ {R}^{3} \epsilon_{0}}$

$\displaystyle\implies \: E \times 4\pi {r}^{2} = \dfrac{ q_{0} \times {r}^{3} }{ {R}^{3} \epsilon_{0}}$

$\boxed{ \displaystyle\implies \: E = \dfrac{ q_{0}r }{ 4\pi \epsilon_{0} {R}^{3} } }$

At r = R :

$\displaystyle\implies \: E = \dfrac{ q_{0}R}{ 4\pi \epsilon_{0} {R}^{3} }$

$\boxed{\displaystyle\implies \: E = \dfrac{ q_{0}}{ 4\pi \epsilon_{0} {R}^{2} } }$

At r > R :

$\displaystyle\therefore \: \oint E \times ds = \dfrac{q}{ \epsilon_{0}}$

$\displaystyle\implies \: \oint E \times ds = \dfrac{q_{0}}{ \epsilon_{0}}$

$\displaystyle\implies \: E \times 4\pi {r}^{2} = \dfrac{q_{0}}{ \epsilon_{0}}$

$\boxed{ \displaystyle\implies \: E = \dfrac{q_{0}}{ 4\pi\epsilon_{0} {r}^{2} } }$

Answer 2

Explanation:

To Derive:

Electric field intensity due to a uniformly charged solid sphere.

Solution:

Let radius of sphere be R :

At a distance r < R:

\displaystyle\therefore \: \oint E \times ds = \dfrac{q}{ \epsilon_{0}} ∴∮E×ds=

ϵ

0

q

\displaystyle\implies \: \oint E \times ds = \dfrac{ \bigg \{\dfrac{ q_{0}(\frac{4}{3} \pi {r}^{3} ) }{ \frac{4}{3}\pi {R}^{3} } \bigg \}}{ \epsilon_{0}} ⟹∮E×ds=

ϵ

0

{

3

4

πR

3

q

0

(

3

4

πr

3

)

}

\displaystyle\implies \: E \oint \: ds = \dfrac{ \bigg \{\dfrac{ q_{0}(\frac{4}{3} \pi {r}^{3} ) }{ \frac{4}{3}\pi {R}^{3} } \bigg \}}{ \epsilon_{0}} ⟹E∮ds=

ϵ

0

{

3

4

πR

3

q

0

(

3

4

πr

3

)

}

\displaystyle\implies \: E \oint \: ds = \dfrac{ q_{0} \times {r}^{3} }{ {R}^{3} \epsilon_{0}} ⟹E∮ds=

R

3

ϵ

0

q

0

×r

3

\displaystyle\implies \: E \times 4\pi {r}^{2} = \dfrac{ q_{0} \times {r}^{3} }{ {R}^{3} \epsilon_{0}} ⟹E×4πr

2

=

R

3

ϵ

0

q

0

×r

3

\boxed{ \displaystyle\implies \: E = \dfrac{ q_{0}r }{ 4\pi \epsilon_{0} {R}^{3} } }

⟹E=

4πϵ

0

R

3

q

0

r

At r = R :

\displaystyle\implies \: E = \dfrac{ q_{0}R}{ 4\pi \epsilon_{0} {R}^{3} } ⟹E=

4πϵ

0

R

3

q

0

R

\boxed{\displaystyle\implies \: E = \dfrac{ q_{0}}{ 4\pi \epsilon_{0} {R}^{2} } }

⟹E=

4πϵ

0

R

2

q

0

At r > R :

\displaystyle\therefore \: \oint E \times ds = \dfrac{q}{ \epsilon_{0}} ∴∮E×ds=

ϵ

0

q

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

ϵ

0

q

0

\displaystyle\implies \: E \times 4\pi {r}^{2} = \dfrac{q_{0}}{ \epsilon_{0}} ⟹E×4πr

2

=

ϵ

0

q

0

\boxed{ \displaystyle\implies \: E = \dfrac{q_{0}}{ 4\pi\epsilon_{0} {r}^{2} } }

⟹E=

4πϵ

0

r

2

q

0