91影视

A parallel plate capacitor is an arrangement of two metal plates connected in parallel separated from each other by some distance. A dielectric medium occupies the gap between the plates.

Given We are given the capacitance of parallel-plate capacitor

Required We are asked to calculate

(a) The charge Q on each plate. and it is connected to battery

(b) The charge Q when the distance d is doubled and the voltage is constant

(c) The charge Q when the radius r is doubled where the voltage and the separated distance are constant.

Solution (a) The capacitor is charged with a battery 12.0 V, this means the potential difference between the two parallel plates is Where the potential difference is related to the capacitance and the charge on the plates by equation 24.2

as next

Now we can plug our values for C and Vab into equation (1 ) to get ttE magnitude of the charge Q on the each plate

Therefore the charge on each plate is

(b) The capacitance is related to ttE separated distance as shown by equation 242

As shown by equation (2

So as the distance doubled this means the capacitance is halved and becomes C = 5.0 PF. Also, as shown by equation (1 ttE capacitance is directly proportional to the charge Q, so if the voltage Vab is constant and the capacitance is halved, therefore the charge Q is halved.

Therefore if separation is doubled then the charge will be

(c) As shown by equation (2), the capacitance depends directly on the area A, where A equals _ Therefore

, 2 So, as the radius is doubled, the capacitance will increase four times and becomes

Now let us plug our values for C and Vab into equation (1) to get Q where Vab is constant

Therefore the charge on the plates if the capacitor were connected to the battery radius of each plate was doubled without changing their separation is