a.)Two 14-cm-diameter electrodes 0.50cm apart form a parallel-plate capacitor. T

Question

a.)Two 14-cm-diameter electrodes 0.50cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 20V battery. After a long time, the capacitor is disconnected from the battery but is not discharged.

b.) What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes right after the battery is disconnected?

c.)What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.2cm apart?

d.) What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after the original electrodes (not the modified electrodes of part b) are expanded until they are 23cm in diameter?

Explanation / Answer

a) q1 = C*v
where C = ?o*A/d = 8.854x10^-12*?*r^2/d = 8.854x10^-12*?*(0.055)^2/0.0057 = 1.476x10^-11F

So q1 = 1.476x10^-11*15 = 2.21x10^-10C and q2 = -2.21x10^-10C

b) E = ?/?o = 2.21x10^-10/(8.854x10^-12*?*r^2) = 2.21x10^-10/(8.854x10^-12*?*(0.055)^2)
= 2630V/m

c) V = E*d = 2630V/m*0.0057 = 15.0V

d) Now C decreases to
C = ?o*A/d = 8.854x10^-12*?*r^2/d = 8.854x10^-12*?*(0.055)^2/0.014 = 6.01x10^-12F

q1 = 2.21x10^-10C and q2 = - 2.21x10^-10 Charge remains constant in this situation

e) E = = 2630V/m It remains the same

f) V = E*d =2630*0.014) = 36.8V

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