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Question

(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.

(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.

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Solution

(i)

Consider a thin infinite uniformly charged plane sheet having the surface charge density of σ. The electric field is normally outward to the plane sheet and is same in magnitude but opposite in direction.

Now, draw a Gaussian surface in the form of cylinder around an axis. Let its cross-sectional area be A.

The cylinder is made from three surfaces A,S2, and A' and the electric flux linked with S2 is 0.So, the total electric flux linked through the Gaussian surface is

ϕE=electric flux through A+electric flux through S2+electric flux through A'

ϕE=EA cos 0∘+EA cos 0∘.

ϕ=2EA ...(i)

According to Gauss theorem,

ϕ=qε0

ϕ=σA2ε0 (∵q=σA) ...(ii)

From equation (i) and (ii)

2EA=σAϵ0

E=σ2ϵ0

The direction of field for positive charge density is in outward direction and perpendicular to the plane infinite sheet whereas for the negative charge density the direction becomes inward and perpendicular to the sheet.

(ii) Given, C1:C2=1:2

⇒C2=2C1

For parallel combination of capacitor,

Cp=C1+C2

=C1+2C1=3C1

The energy stored in capacitor

E=12CpV2p

=123C1V2p=32C1V2p ....(i)

For series combination of capacitor,

1Cs=1C1+1C2

Cs=23C1

The energy stored in capacitor

E=12CsV2s

E=C1V2s3 ...(ii)

Equating equation (i) and (ii),we get

32C1V2p=C1V2s3

VpVs=√23

Consider a thin infinite uniformly charged plane sheet having the surface charge density of σ. The electric field is normally outward to the plane sheet and is same in magnitude but opposite in direction.

Now, draw a Gaussian surface in the form of cylinder around an axis. Let its cross-sectional area be A.

The cylinder is made from three surfaces A,S2, and A' and the electric flux linked with S2 is 0.So, the total electric flux linked through the Gaussian surface is

ϕE=electric flux through A+electric flux through S2+electric flux through A'

ϕE=EA cos 0∘+EA cos 0∘.

ϕ=2EA ...(i)

According to Gauss theorem,

ϕ=qε0

ϕ=σA2ε0 (∵q=σA) ...(ii)

From equation (i) and (ii)

2EA=σAϵ0

E=σ2ϵ0

The direction of field for positive charge density is in outward direction and perpendicular to the plane infinite sheet whereas for the negative charge density the direction becomes inward and perpendicular to the sheet.

(ii) Given, C1:C2=1:2

⇒C2=2C1

For parallel combination of capacitor,

Cp=C1+C2

=C1+2C1=3C1

The energy stored in capacitor

E=12CpV2p

=123C1V2p=32C1V2p ....(i)

For series combination of capacitor,

1Cs=1C1+1C2

Cs=23C1

The energy stored in capacitor

E=12CsV2s

E=C1V2s3 ...(ii)

Equating equation (i) and (ii),we get

32C1V2p=C1V2s3

VpVs=√23

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