Question

Derive the expression for the electric field at the surface of a charged conductor

Answer 1

Electric field at the surface of charged conductor is given by
Er = σ/∈₀[r/3 - r²/4R]
where σ is surface charge density , R is radius of sphere and r ≤ R
Now, for finding maximum Electric field , we have to differentiate Er with respect to r .
e.g., dEr/dr = σ/∈₀[1/3 - 2r/4R ]
When dEr/dr = 0 , r = 2R/3
Now, check d²Er/dr² values at r = 2R/3 you will get d²Er/dr² < 0
Means , at r = 2R/3 , Er will be maximum.
∴ Emax = σ/∈₀[2R/3×3 - 4R²/9×4R]
= σ/∈₀[ 2R/9 - R/9]
= σR/9∈₀

Answer 2

Answer:

Explanation:

Electric field at the surface of charged conductor is given by

Er = σ/∈₀[r/3 - r²/4R]

where σ is surface charge density , R is radius of sphere and r ≤ R

Now, for finding maximum Electric field , we have to differentiate Er with respect to r .

e.g., dEr/dr = σ/∈₀[1/3 - 2r/4R ]

When dEr/dr = 0 , r = 2R/3

Now, check d²Er/dr² values at r = 2R/3 you will get d²Er/dr² < 0

Means , at r = 2R/3 , Er will be maximum.

∴ Emax = σ/∈₀[2R/3×3 - 4R²/9×4R]

= σ/∈₀[ 2R/9 - R/9]

= σR/9∈₀