two 10 cm diameter electrodes .5 cm apart form a parallel plate capacitor. The e

Question

two 10 cm diameter electrodes .5 cm apart form a parallel plate capacitor. The electrodes are attached by metal wires to the terminals of a 15V battery. What are the charge on each electrode, the electric field strength inside the capacitor and the potential difference between the electrodes:

(a) while the capacitor is attached to the battery?

(b) after insulating the handles are use to pull the electrodes away from each other until they are 1.0 cm apart? The electrodes remain connected to the battery during this process.

Explanation / Answer

a) q1 = C*v
where C = ?o*A/d = 8.854x10^-12*?*r^2/d = 8.854x10^-12*?*(0.05)^2/0.005 = 1.391x10^-11F

So q1 = 1.391x10^-11*15 = 2.09x10^-10C and q2 = -2.09x10^-10C

E = ?/?o = 2.09x10^-10/(8.854x10^-12*?*r^2) = 2.21x10^-10/(8.854x10^-12*?*(0.05)^2)
= 3000V/m

V = E*d = 3000V/m*0.005 = 15.0V


After insulating

b) Now C decreases to
C = ?o*A/d = 8.854x10^-12*?*r^2/d = 8.854x10^-12*?*(0.05)^2/0.01 = 6.954x10^-12F

q1 = 2.09x10^-10C and q2 = - 2.09x10^-10 Charge remains constant in this situation

E = 3000V/m It remains the same

V = E*d =3000*0.01 = 30V

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