b) In the figure there are three infinite long thin sheets having surface
charge density +2σ, -2σ and +σ respectively. Give the magnitude and
direction of electric field at a point to the left of sheet of charge density
+2σ and to the right of sheet of charge density +σ.
SAM-PLE PAPER PHYSICS 2021 Q-31
Answers
Explanation:
NSWER
(a) The charge of magnitude +q present inside the spherical shell induces a charge of magnitude −q 0n the inner part of the shell. Therefore, for the charge on the inner part, the surface charge density will be due to −q.
It is given by:
σ
1
=
surface area of inner part
Totalcharge
⇒
4πr
1
2
−q
The charge on the outer shell will be +q and the surface charge density at the outer part will be due to sum of q and Q charges present on the outer shell.
σ
2
=
Outersurfacearea
Totalcharge
=
4πr
2
2
Q+q
...(ii)
(b) If we consider a loop inside an irregular conductor, then will be no work done on it as there is no field present inside the cavity. So, there will be no field inside the cavity even if it is irregular in sha
Given:
Three infinitely long sheets with surface charge densities +2σ, -2σ and +σ respectively
To find:
Magnitude and the direction of the electric field at a point to the left of the sheet of charge density +2σ and to the right of the sheet of charge density +σ.
Solution:
- From Gauss' law
dϕ = ∫ E . ds = q / ε
From here,we get electric field E as
E = q / (ε × ds)
- We know that Surface charge density σ is
σ = q / ds
- So , E = σ / ε
- As per the figure attached;
= 2σ / 2ε (-i) + 2σ / 2ε (i) + σ / 2ε (-i)
= σ / 2ε (-i)
= σ / 2ε (i) + 2σ / 2ε (-i) + 2σ / 2ε (i)
