A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x=3d. The slab is equidistant from the plates. The capacitor is given some charge. As x goes from 0 to 3d :
Explanation
Three charges Q, +q and +q are placed at the vertices of a right angled isosceles triangle as shown in the figure. The net electrostatic energy of the configuration is zero if Q is equal to :
An alpha particle of 5MeV at a large distance proceeds towards a gold nucleus (Z=79) to make a head on collision. The closest distance of approach from the centre of gold nucleus is:
Using
Energy=12×Cequivalent×V2
We need to find the equivalent capacitance of the capacitor.
Using law for series combination of capacitors,
1Cnet=1C1+1C2+1C3
where, C1=Aϵ0x ( x is the width between capacitor walls and dielectric on one side )
C2=Aϵ0κ(d2)=4Aϵ0d
C3=Aϵ0(d2−x)
therefore,
1Cnet=x(Aϵ0)+(d2−x)(Aϵ0)+d(4Aϵ0)=3d(4Aϵ0)⇒Cnet=4Aϵ03d
Therefore energy =2Aϵ0V23d
The capacitance of a parallel plate capacitor is given by
C=εoεrAd,
where A is the area of the plate and d is the distance between them.
In the first case, there was no dielectric medium, i.e., εr=1.
So we have C=εoAd
In the second case, the distance is doubled, i.e., 2d and a substance of dielectric constant 3 is inserted between the plates.
Hence capacitance will change to
C′=3εoA2d
So, C′C=32
⟹C′=32C
For option (B) solution : In XWY, charge on capacitor plate =KCE
In XYW, charge on capacitor plate = CE
For option (C) solution : In WXY, U = 12KCE2
In XYW, U = CE22K
For option (D) solution : 2 both cases, electric field = Ed .
Potential for each plate remain same over whole are if potential difference thems is way V then V = Ed i.e. E is also same inside the plates
To Keep E same free change dansity is changed i.e. charge redistributes itself
To find new capacitance, two capacitors can be taken as connected in parallel Then
Ceq=5.ϵ0A/3d+ϵ0.2A/3d=7ϵ0A3d
Q = CV, as Q remains unchanged V is changed to 37V
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