Q. 26

# (i) Use Gauss's l

(i) To find the electric field due to infinite uniform charged plane sheet we take a cylindrical Gaussian surface such that the surface passes perpendicular to the plane of the sheet. The surface charge density of the sheet is taken as σ . The electric field is normally outward to the plane sheet and is of same magnitude in both the directions. The cross sectional area of the Gaussian surface is taken to be A. The Gaussian cylinder consists of three surfaces P, S2 and P’ but the electric flux because of the sheet linked with the surface S2 is 0.

Hence, the total flux linked through the Gaussian surface is :

ΦT = electric flux through S2 + electric flux through P’ + electric flux through P

ΦT = 0 + EA cos 0° + EA cos 0°

ΦT = 2 EA …….(i)

The gauss theorem states that: Where,

‘q’ is the charge is the charge on the plate given as, q = σA

ϕ is the flux through the Gaussian surface, is the permittivity in air. ……..(ii)

Equating equation (i) and (ii) we get,  This is the electric field due to uniformly charged infinite plane sheet.

The direction of field is perpendicular to the plane of the sheet and points outward in the case of positive charge density. Whereas, in the case of negative charge it is in the opposite direction .i.e. perpendicularly inward.

(i) When the capacitors are connected in parallel their combination capacitance comes out to be,

C = C1 + C2

When they are connected in series their combination capacitance comes out to be, The energy stored in capacitor is: Where,

C is the total combination capacitance,

V is the potential difference across the capacitors

Given,

Relation between two capacitances,

C2 = 2C1

Thus, the total capacitance in parallel combination becomes:

C = C1 + 2C1 = 3C1

Thus, the total energy stored in parallel combinationis: E = C1V2 ……(i)

Where,

V is the potential across the parallel capacitors

Total capacitance for series combination becomes: C’ = C1

Energy stored in series combination:

E = C’V’2

E = C1V’2 ……..(ii)

Where,

V’ is the potential across the series capacitor combination.

Equation both equation (i) and (ii). We get: C1V2 = C1V’2  This is the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2.

OR

(i) (a) We suppose a parallel plate capacitor made up of two identical plates P and Q having an area “A” each and separated by a distance ‘d’.

The space between the plates is filled by a dielectric medium with the value of dielectric constant K and a surface charge density σ on each plates On plate I the surface charge density is given as:

σ = Q / A

Where,

Q is the total charge on the plate,

A is the surface area of the plate.

The surface charge density on plate II is -σ .

Electric field in the outer region of plate I is given as: Electric field in the outer region of plate II is given as: The inner region between the two plates have electric fields that add up to give:  This is the electric field in the inner region of a parallel plate capacitor which is in the direction from positively charged plate to negatively charged.

(b) In the case of uniform electric field , The relation between the potential difference and the electric field is given as:

E = V / d

Where,

V is the potential difference between the plates,

‘d’ is the separation between the plates

Therefore,

V = Ed = …..(a)

The potential difference is given as (c) The capacitance so formed is given as,

C = Replacing the value of V from (a) we get, The capacitance so formed is given as (d) The potential difference of a metallic sphere is given as: Where,

R is the radius of the sphere,

Q is the charge on the sphere,

Given,

The radius of the metallic sphere = R,

The surface charge density = σ

The potential then is:

VR = VR = …..(i)

Similarly,

Given,

Radius of metallic sphere = 2R

The surface charge density = σ

the potential of a metallic sphere is :

V2R = V2R = …….(ii)

Comparing equation (i) and (ii) , we know that V2R> VR, therefore a potential difference is produced and the charge will flow from higher potential to lower potential, i.e. from the metallic sphere of radius 2R to radius R , if they are connected.

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