Now the friends try a homework problem.A capacitor is constructed with two parallel metal plates each with an area of 0.74 m2 and separated by d = 0.80 cm. The two plates are connected to a 4.0-volt battery. The current continues until a charge of magnitude Q accumulates on each of the oppositely charged plates.Find the electric field in the region between the two plates. V/mFind the charge Q. CFind the capacitance of the parallel plates.
Question
Now the friends try a homework problem.A capacitor is constructed with two parallel metal plates each with an area of 0.74 m2 and separated by d = 0.80 cm. The two plates are connected to a 4.0-volt battery. The current continues until a charge of magnitude Q accumulates on each of the oppositely charged plates.Find the electric field in the region between the two plates. V/mFind the charge Q. CFind the capacitance of the parallel plates.
Solution
Sure, let's solve this step by step.
- Find the electric field in the region between the two plates.
The electric field E between the plates of a capacitor is given by the equation E = V/d, where V is the voltage and d is the distance between the plates. Here, V = 4.0 volts and d = 0.80 cm = 0.008 m. So,
E = V/d = 4.0 volts / 0.008 m = 500 V/m.
- Find the charge Q.
The electric field E is also related to the charge Q on the capacitor by the equation E = Q/(ε0A), where ε0 is the permittivity of free space (ε0 = 8.85 x 10^-12 C^2/Nm^2) and A is the area of one of the plates. Here, A = 0.74 m^2. So,
Q = Eε0A = 500 V/m * 8.85 x 10^-12 C^2/N*m^2 * 0.74 m^2 = 3.28 x 10^-9 C.
- Find the capacitance of the parallel plates.
The capacitance C of a capacitor is given by the equation C = Q/V, where Q is the charge on the capacitor and V is the voltage across the plates. Here, Q = 3.28 x 10^-9 C and V = 4.0 volts. So,
C = Q/V = 3.28 x 10^-9 C / 4.0 volts = 8.2 x 10^-10 F.
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