Electric field intensity due to a uniformly charged solid sphere
Answers
To Derive:
Electric field intensity due to a uniformly charged solid sphere.
Solution:
- Let radius of sphere be R :
At a distance r < R:
At r = R :
At r > R :
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