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CBSE Class 12 Physics 2016 Outside Delhi Set 1 Paper

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Question : 26
Total: 26
(i) Use Gauss's law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities?
(ii) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances in the ratio 1:2 so that the energy stored in the two cases becomes the same.
OR
(i) If two similar large plates, each of area A having surface charge densities +σ and −σ are separated by a distance d in air, find the expressions for
(a) field at points between the two plates and on outer side of the plates.
Specify the direction of the field in each case.
(b) the potential difference between the plates.
(c) the capacitance of the capacitor so formed.
(ii) Two metallic spheres of radii R and 2R are charged so that both of these have same surface charge density σ. If they are connected to each other with a conducting wire, in which direction will the charge flow and why?
Solution:  
(i) Derivation for electric field due to infinite plane Sheet of charge Directions of field
(ii) Formula
Calculation and result
Symmetry of situation suggests that
→
E
 is perpendicular to the plane. A Gaussian surface is considered through P like a cylinder of flat caps parallel to the plane and one cap passing through P. The plane being the plane of symmetry for the Gaussian surface.

∮
→
E
⋅
→
ds
=
∫
→
E
⋅
→
ds
‌through caps ‌
→
E
⊥
→
ds
 for all over curved surface and hence
→
E
⋅
→
ds
=0
∫
caps
Eds‌=2E∆s
∆s‌=‌ area of each cap ‌
By Gauss' law
∮
→
E
⋅
→
ds
‌=‌
q
ε0
=‌
σ∆s
ε0
∴ 2E∆s‌=‌
σ∆s
ε0
E‌=‌
σ
2ε0
If σ is positive
→
E
 points normally outwards/ away from the sheet
If σ is (−) ve
→
E
 points normally inwards/towards the sheet
Us‌=‌
1
2
CsVs2
Up‌=‌
1
2
CpVp2
⇒‌‌‌
V‌series ‌
V‌parallel ‌
‌=√‌
C‌equivalent parallel ‌
C‌equivalent series ‌
‌=√‌
C1+C2
C1C2
C1+C2
‌=‌
C1+C2
√C1C2
=‌
3
√2
OR(i) Deriving the expression for field between the plate & outside
Direction of electric field inside and outside
Potential difference between the plates
Capacitance
(ii) Direction of flow of charge

Inside
→
E
‌=
→
E
1+
→
E
2
‌=‌
σ+σ
2E0
=‌
σ
E0
Outside
→
E
‌=
→
E
2−
→
E
1
‌=‌
σ−σ
2ε0
=0
(b) Potential difference between plates
V=Ed=‌
1
ε0
‌
Qd
A
(c) Capacitance
C=‌
Q
V
=‌
ε0A
d
(ii) As potential on and inside a charged sphere is given
V=‌
1
4πε0
‌
q
r
=‌
1
4πε0
⋅‌
4πr2σ
r
∴V‌‌∝r
Hence, the bigger sphere will be at higher potential, so charge will flow from bigger sphere to smaller sphere.
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