Two 13.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. T

Question

Two 13.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 13V battery.

A. What is the charge on each electrode while the capacitor is attached to the battery?

B. What is the electric field strength inside the capacitor while the capacitor is attached to the battery?

C. What is the potential difference between the electrodes while the capacitor is attached to the battery?

D. What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.4 cm apart? The electrodes remain connected to the battery during this process.

E. What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each other until they are 1.4 cm apart? The electrodes remain connected to the battery during this process

F. What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.4 cm apart? The electrodes remain connected to the battery during this process.

Explanation / Answer

a)

The capacitance can be calculated from:

C = *A/d

Calculate the Area:

A = *r^2 = *(.13/2)^2 = 13*10^-3 m^2

d = .50/100 = 5*10^-3 m

C = 8.854*10^-12 *13*10^-3 / 5*10^-3 = 2.3*10^-11 = .23 pF

Q = C*V = 2.3*10^-11 * 13 = 2.99*10^-10 C

b)

The electric field is the voltage divided by the separation distance:

E = V/d = 13/5*10^-3 = 2.6*10^3 V/m

c)

The potential difference is simply the battery voltage, since the voltage will remain constant immediately after battery disconnection:

V = 13 V

d) Now C decreases to
C = o*A/d = 8.854x10^-12**r^2/d = 8.854x10^-12**(0.065)^2/0.014 =?

Charge remains constant in this situation
e. 2.6*10^3 It remains the same

f) V = E*d =2600*0.014) = 36.4V

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