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Electric Charges And Fields

Question
CBSEENPH12039360

(i) If two similar large plates, each of area A having surface charge densities 1s and 2s 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 s. If they are connected to each other with a conducting wire, in which direction will the charge flow and why ?

Solution
i)
  

Given, each plate has an area A and surface charge densities 1s and 2s respectively.
The plates 1 and 2 be separated by a small distance d.
For plate 1:
Surface charge density, straight sigma space equals space straight Q over straight A
For plate 2:
Surface charge density, straight sigma space equals space fraction numerator negative straight Q over denominator straight A end fraction equals negative straight sigma
Electric field in different regions:
Outside region 1,
E = fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction minus fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction equals 0
Outside region 2,
straight E equals space space fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction minus fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction equals 0
Inner region:
In the inner region between plates 1 and 2, electric fields due to the two charged plates add up.
That is,
straight E space equals space fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction plus space fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction
space space space space equals straight sigma over straight epsilon subscript straight o
space space space space equals space fraction numerator straight Q over denominator straight epsilon subscript straight o straight A end fraction
Direction of electric field is from positive to the negative plate.
b) Potential difference between the plates, V = Ed = 1 over straight epsilon subscript straight o fraction numerator Q d over denominator A end fraction
c) Capacitance of the capacitor so formed, C = straight Q over straight V equals fraction numerator epsilon subscript o A over denominator d end fraction
ii) Potential on and inside of charged sphere is given by,
straight V space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction. straight q over straight r

space space space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction. fraction numerator 4 πr squared space straight sigma over denominator straight r end fraction

therefore space straight V space proportional to space straight r
That is, the sphere having larger radius will be at a higher potential.
Therefore, charge will flow from bigger sphere to smaller sphere.

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