Final answer:
A) The potential difference before the Mylar is inserted is 3.00 V. B) The electric field before the Mylar is inserted is 30.0 V/mm. C) The charge before the Mylar is inserted is 9.00 μC. D) The capacitance after the Mylar is inserted is 1.20 μF. E) The electric field after the Mylar is inserted is 75.0 V/mm. F) The charge after the Mylar is inserted is 9.00 μC.
Step-by-step explanation:
A) Before the Mylar is inserted, the potential difference between the plates of the capacitor can be found using the formula:
V = Q/C,
where V is the potential difference, Q is the charge, and C is the capacitance. Since the charge on each plate is 9.00 μC and the capacitance is 3.00 μF, we can substitute these values into the formula:
V = (9.00 μC)/(3.00 μF) = 3.00 V.
Therefore, the potential difference before the Mylar is inserted is 3.00 V.
B) The electric field between the plates can be found using the formula:
E = V/d,
where E is the electric field, V is the potential difference, and d is the distance between the plates. Substituting the values, we get:
E = (3.00 V)/(0.10 mm) = 30.0 V/mm.
Therefore, the electric field before the Mylar is inserted is 30.0 V/mm.
C) The charge on the capacitor can be found using the formula:
Q = C*V,
where Q is the charge, C is the capacitance, and V is the potential difference. Substituting the values, we get:
Q = (3.00 μF)*(3.00 V) = 9.00 μC.
Therefore, the charge before the Mylar is inserted is 9.00 μC.
D) After the Mylar is inserted, the capacitance of the capacitor will change. However, the charge on the plates will remain constant, so we can use the formula:
V = Q/C,
where V is the potential difference, Q is the charge, and C is the capacitance. Rearranging the formula, we get:
C = Q/V.
Substituting the values, we get:
C = (9.00 μC)/(7.5 V) = 1.20 μF.
Therefore, the capacitance after the Mylar is inserted is 1.20 μF.
E) The electric field between the plates can be found using the formula:
E = V/d,
where E is the electric field, V is the potential difference, and d is the distance between the plates. Substituting the values, we get:
E = (7.5 V)/(0.10 mm) = 75.0 V/mm.
Therefore, the electric field after the Mylar is inserted is 75.0 V/mm.
F) The charge on the capacitor remains constant, so the charge after the Mylar is inserted is still 9.00 μC.