Two capacitors are identical, except that one is empty and the other is filled with a dielectric (k = 3.8). The empty capacitor is connected to a 16 -V battery. What must be the potential difference across the plates of the capacitor filled with a dielectric so that it stores the same amount of electrical energy as the empty capacitor?
To find the potential difference across the plates of the capacitor filled with a dielectric, we need to consider the relationship between capacitance, charge, and voltage.
The electrical energy stored in a capacitor can be calculated using the formula:
E = (1/2) * C * V^2
where E is the electrical energy, C is the capacitance, and V is the potential difference (voltage) across the capacitor plates.
Since the two capacitors are identical except for the dielectric in one of them, we can assume that their capacitances are the same. Let's denote the capacitance of each capacitor as C.
For the empty capacitor, we know that the potential difference across the plates is 16 V, so the energy stored is:
E1 = (1/2) * C * (16 V)^2
For the capacitor filled with a dielectric, let's assume the potential difference across its plates is V2. Now, the energy stored is:
E2 = (1/2) * C * (V2)^2
To find the potential difference V2, we equate the energies E1 and E2:
E1 = E2
(1/2) * C * (16 V)^2 = (1/2) * C * (V2)^2
Now we can cancel out the common factors and solve for V2:
(16 V)^2 = (V2)^2
256 V^2 = V2^2
Taking the square root of both sides, we get:
V2 = 16 V
Therefore, the potential difference across the capacitor plates filled with the dielectric needs to be 16 V in order to store the same amount of electrical energy as the empty capacitor.