Question

Two 10.0-cm-diameter electrodes 0.59 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14 Vbattery.

What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.1 cm apart? The electrodes remain connected to the battery during this process.

q1, q2= ____ C

Note: Answer is not 8.9*10^{-10},-8.9*10^{-10} C

Answer 1

Area of the disk = d²/4
where d is the diameter of the disk = 10 cm
Area = 0.00785 m²
We know that the capacitance of parallel plate capacitor = A /d
where d is the distance between them.
After pulling away the distance is = 1.1 cm = 1.1*10^-2 m
Now putting the values for capacitance
C = 8.85*10^-12*0.00785 /(1.1*10^-2) = 6.316*10^-12 F
We know that the charge on the plate
Q = CV
where V is the voltage of the battery
Q = 6.316*10^-12 *14 = 8.841*10^-11 C
Hence charge on the upper plate
q₁ = 8.841*10^-11 C
and on the lower plate
q₂ = - 8.841*10^-11 C