Page 20 - English Class X_cbse new (FINAL).cdr

S. Solution Marking S. Solution Marking
No. Scheme No. Scheme

24. E = 2π q r2 25. Now, Gaussian surface is outside the given charged shell, so charge
∈0 cont. enclosed by Gaussian surface is Q. Hence, by Gauss’s theorem
cont.

∴ The charge density σ = q ⇒ q = πr2σ ∫= E0 idS = 1 × chargedenclosed
S s ε0
[Where S–area of circle]

E = πr 2 σ [1] ⇒ E0 4πr2 = 1 ×Q ⇒ E0 = 1 Q
2π ∈0 r2 ε0 4πε0 r2
1
E = σ , vectorically E = σ n  2 Thus,electric field outside a charged thin spherical shell is the same
2 ∈0 2 ∈0 as if the whole charge Q is concentrated at the centre.
[3]
where n is a unit vector normal to the plane and going away from it. If σ is the surface charge density of the spherical shell, then
When σ > 0, E is directed away from both sides. Hence electric field Q = 4πR2 σ coulomb

intensity is independent of r. ∴ E0 = 1 4πR2σ = R2σ
Note: For conducting sheet, the surface charge density on both the 4πε0 r2 ε0r2
surface of sheet will be same.
(ii) Electric field inside the shell (hollow charged conducting sphere):
∴E= σ The charge resides on the surface of a conductor. Thus a hollow
∈0 charged conductor is equivalent to a charged spherical shell.-
To find the electric field inside the shell, we consider a spherical
25. (i) Electric field intensity at a point outside a uniformly charged thin Gaussian surface of radius r(<R),
spherical shell: Consider a uniformly charged thin spherical shell
of radius R carrying charge Q. To find the electric field outside the [1]
shell, we consider a spherical Gaussian surface of radiusr (>R),
[1] concentric with the given shell. If E is the electric field inside
concentric with given shell. If E is the shell, then by symmetry electric field strength has the same
electric field outside the magnitude Ei on the Gaussian surface and is directed radially
outward. Also the directions of normal at each point is radially
shell, then by symmetry electric
field strength has same outward, so angle between Ei and dS is zero at each point.

magnitude E0 on the Gaussian Hence, electric flux through Gaussian surface
surface and is directed radially

outward. Also the directions of
normal at each point is radially

outward, so angle between Ei and

dS is zero at each point.

Hence, electric flux through Gaussian surface E0 idS .

∫S

∫ = E0dS cos0 = E0 ⋅ 4πr2

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