Two 14.0 -diameter electrodes 0.59 apart form a parallel-plate capacitor. The el

Question

Two 14.0 -diameter electrodes 0.59 apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 12 battery. After a long time, the capacitor is disconnected from the battery but is not discharged. What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.7 apart? --> For this one, I have tried C=(EoA)/d, then Q=C*12, and I got the answer 9.6E-11, but mastering physics says this is wrong. The answers for the following two problems depend on this answer, so I haven't been able to do them. 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 1.7 apart? What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.7 apart?

Explanation / Answer

a) q1 = C*v
where C = ?o*A/d = 8.854x10^-12*?*r^2/d = 8.854x10^-12*?*(0.07)^2/0.0059 = 2.31x10^-11 F

So q1 = 2.31x10^-11*12 = 2.21x10^-10C and q2 = -2.7721x10^-10C

Now C decreases to
C = ?o*A/d = 8.854x10^-12*?*r^2/d = 8.854x10^-12*?*(0.07)^2/0.017 = 8.017455x10^-12 F

q1 = 2.7721x10^-10C and q2 = - 2.7721x10^-10 Charge remains constant in this situation

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