The parallel plates of a capacitor have an area `0.2 m^2` and are `10^(–2)` m apart. The original potential difference between them is 3000 V, and it decreases to 1000 V when a sheet of dielectric is inserted between the plates filling the full space. Compute :
` ( in _0 =9xx 10 ^(-12) S.I ` units )
Permittivity of the dielectric.

The parallel plate of a capacitor have an area 0.2 m^(2) and are 10^(-2) m apart. The original potential difference between them is 3000V, and it decreases to 1000 when a sheet of dielectric is inserted between the plates. Compute (a) Original capacitance C_(0) (b) The original charge Q on each plate . (C) Capacitance C after insertion of the dielectric . (d) Dielectric constant K . (e) The original field E_(0) between the plates and . (f) The electric field E_(0) between the plates and . (f) The electric field E after insertion of the dielectric . (epsi_(0) = 8.85 xx 10^(-12) S.I unit).

The plates of a parallel plate capacitor are 5mm apart and 2m^2 in area. The plates are in vacuum. A potential difference of 10000 V is applied across the capacitor. Compute The electric field in the space between them.

An uncharged parallel plate capacitor is connected to a battery. The electric field between the plates is 10V/m. Now a dielectric of dielectric constant 2 is inserted between the plates filling the entries space. The electric field between the plates now is

An uncharged parallel plate capacitor is connected to a battery. The electric field between the plates is 10V/m. Now a dielectric of dielectric constant 2 is inserted between the plates filling the entries space. The electric field between the plates now is

The plates of a parallel plate capacitor are 5mm apart and 2m^2 in area. The plates are in vacuum. A potential difference of 10000 V is applied across the capacitor. Compute The charge on each plate

The area of parallel plates of an air capacitor is 0.2 m^(2) and the distance between them is 0.01m The potential difference between the plates, the potential difference between the plates is 3000 V . When a 0.01 m thick sheet of an insulating material is placed between the plates, the potential difference decrease to 1000 volt. Determine (i) capacitance of capacitance before placing the sheet (ii) charge on each plate (iii) dielectric constant of material (iv) capacitanc after placing the insulator (v) absoulate permittivity of the dielectric.

The area of parallel plates of an air capacitor is 0.2 m^(2) and the distance between them is 0.01m The potential difference between the plates, the potential difference between the plates is 3000 V . When a 0.01 m thick sheet of an insulating material is placed between the plates, the potential difference decrease to 1000 volt. Determine (i) capacitance of capacitance before placing the sheet (ii) charge on each plate (iii) dielectric constant of material (iv) capacitanc after placing the insulator (v) absoulate permittivity of the dielectric.

Two plates are 2 cm apart. If a potential difference of 10 V is applied between them. The electric field between the plates will be

A capacitor with air as the delectric is charged to a potential of 100 volts. If the space between the plates is now filled with a dielectric of dielectric constant 10 , the potential difference between the plates will be