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 Q₀ on its plates.The plates are pulled apart to a separation 2d while the capacitor remains connected to the battery.After the plates are 2d apart,the magnitude of the charge on the plates and the potential difference between them are A) $\frac { 1 } { 2 }$ Q₀, $\frac { 1 } { 2 }$ V₀ B) $\frac { 1 } { 2 }$ Q₀,V₀ C)Q₀,V₀ D)2Q₀,V₀ E)2Q₀,2V₀

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Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor is given by, C = εA/d, where ε is the permittivity of free space, A is the area of the plates, and d is the separation between them. Since the plates are pulled apart to a separation of 2d, the capacitance becomes Cs = εA/(2d). Since the capacitor is still connected to the battery, the potential difference between the plates remains the same, i.e., Vs = V₀. We know that Q = CV, where Q is the charge on the plates and V is the potential difference between them. So, the charge on the plates after the separation is 2d becomes Qs = C * Vs = Cs * Vs = (εA/(2d)) * V₀ = 1/2 * Q₀. Therefore, the answer is option B) 11ea99f6_d70a_6a62_8bb6_71b8cff4825d_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11 Q₀,V₀.

A Parallel Plate Capacitor of Capacitance C0 Has Plates of Area

A Parallel Plate Capacitor of Capacitance C₀ Has Plates of Area