Find the capacitance between A and B if ...

Find the capacitance between A and B if two dielectric alabs (each of area `A`) of dielectric constants `K_(1)` and `K_(2)` and thicknesses `d_(1)` and `d_(2)` are inserted between the plates of a parallel plate capacitor of plate area `A`.
.

`C=Aepsilon_(0)[(K_(1)K_(2))/(K_(2)d_(1)+K_(1)d_(1))]`

`C=2Aepsilon_(0)[(K_(1)K_(2))/(K_(2)d_(1)+K_(1)d_(2))]`

`C=Aepsilon_(0)[(K_(1)K_(2))/(2K_(2)d_(1)+K_(1)d_(2))]`

`C=Aepsilon_(0)[(K_(1)K_(2))/(K_(2)d_(1)+K_(1)d_(2))]`

Text Solution

Verified by Experts

The correct Answer is:

The combination is equivalent to two capacitors connected in series
`1/(C)=1/(C_(1))+1/(C_(2))`
`=d_(1)/(AK_(1)epsilon_(0))+d_(2)/(AK_(2)epsilon_(0))=1/(Aepsilon_(0))[(K_(2)d_(1)+K_(1)d_(2))/(K_(1)K_(2))]`
or `C=Aepsilon_(0)[(K_(1)K_(2))/(K_(2)d_(1)+K_(1)d_(2))]`

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Find the capacitance between A and B if two dielectric slabs (each of thickness d) of dielectric constants K_(1) and K_(2) and areas K_(1) and K_(2) and inserted between the plates of a parallel plate capacitor of plate area A . .

`C_("equivalent")=(K_(1)epsilon_(0)A_(2))/d=(epsilon_(0))/d(K_(1)A_(1)+K_(2)A_(1))`

`C_("equivalent")=(epsilon_(0))/d(K_(2)A_(1)+K_(2)A_(2))`

`C_("equivalent")=(K_(1)epsilon_(0)A_(2))/d=(epsilon_(0))/d(K_(1)A_(1)+K_(2)A_(2))`

`C_("equivalent")=(K_(1)epsilon_(0)A_(2))/d=(epsilon_(0))/d(K_(1)A_(1)+K_(2)2A_(2))`

A dielectric slab is placed between the plates of a parallel plate capacitor. Its capacitance

becomes zero

remains the same

decreases

increases

Two thin dielectric slabs of dielectric constants K_(1) and K_(2) (K_(1) lt K_(2)) are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field E between the plates with distance d as measured from plate P is correctly shown by

Find the capacitance between A and B if two dielectric alabs (each of area `A`) of dielectric constants `K_(1)` and `K_(2)` and thicknesses `d_(1)` and `d_(2)` are inserted between the plates of a parallel plate capacitor of plate area `A`.
.

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