1.

In figure , the plates of a parallel plate capacitors have unequal charges. Its capacitance is C. P is a point outside the capacitor and close to the plate of charge -Q. The distance between the plates is d. Then

Answer»

a point CHARGE at point P will experience electric force due to the capacitors
the potential difference between the plates will be `3Q//2C`.
the energy stored in the electric field in the REGION between the plates is `9Q^2//8C`.
the force on one PLATE due to the other plate is `Q^2//2piepsilon_0d^2`.

Solution :a.,b.,c.
i.`E=(2Q)/(2Aepsilon_(0))+(Q)/(2Aepsilon_(0))=(3Q)/(2Aepsilon_(0))`
or `E=(3)/(2)(Q)/(cd)` or `Ed=(3Q)/(2C)=V`
ii. `F=E(-Q)=((2Q)/(2Aepsilon_(0)))xx((-Q))/(1)=(Q^(2))/(Aepsilon_(0))`
`=(Q^(2))/(Aepsilon_(0))` ltbgt iii. Energy `=(1)/(2)epsilon_(0)E^(2)Ad=(1)/(2)epsilon_(0)((3Q)/(2Cd))^(2)Ad=(9)/(8)(Q^(2))/(C)`


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