A parallel plate capacitor of capacitance C₀ has plates of area A with separation d between them.When it is connected to a battery of voltage V₀,it has charge of magnitude Q₀on its plates.It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3.After the dielectric is added,the magnitudes of the charge on the plates and the potential difference between them are A) $\frac { 1 } { 3 }$ Q₀, $\frac { 1 } { 3 }$ V₀. B)Q₀, $\frac { 1 } { 3 }$ V₀. C)Q₀,V₀. D)Q₀,3V₀. E)3Q₀,3V₀.

Question

Quizplus · Accepted Answer

The capacitance of the capacitor after adding the dielectric is: C' = kε0A/d, where k is the dielectric constant. So, the charge on the plates remains the same: Q' = Q = C₀V₀ And, the potential difference between the plates becomes: V' = Q'/C' = Qd/ε0kA = V₀/3 Therefore, the correct answer is B: Q₀, 11ea99f6_d70d_50be_8bb6_d1de3a01b0d2_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11 V₀.

A Parallel Plate Capacitor of Capacitance C0 Has Plates of Area

A Parallel Plate Capacitor of Capacitance C₀ Has Plates of Area