How does the energy stored in a capacitor change if (a) the potential difference is
doubled, and (b) the charge on each plate is doubled, as the capacitor remains connected to a battery?
Does the stored energy go up by a factor of 4 for both A and B?
a. Energy = 0.5*C*V^2
V doubled:
Energy = 0.5*C*(2V)^2 = 0.5*C*4V^2
So the energy is increased by a factor of 4.
b. Yes.
Yes
(a) If the potential difference across a capacitor is doubled, the energy stored in the capacitor will increase by a factor of 4.
To understand this, we need to know that the energy stored in a capacitor is given by the formula:
E = (1/2) * C * V^2
Where:
E = Energy stored in the capacitor
C = Capacitance of the capacitor
V = Potential difference across the capacitor
In this case, let's assume that the capacitance remains constant. If the potential difference is doubled (V' = 2V), the energy stored can be calculated as:
E' = (1/2) * C * (2V)^2
= (1/2) * C * 4V^2
= 2 * (1/2) * C * V^2
= 2E
From the above calculation, we can see that the energy stored in the capacitor is doubled when the potential difference is doubled.
(b) If the charge on each plate of a capacitor is doubled while the potential difference remains constant, the energy stored in the capacitor will increase by a factor of 2.
Again, using the energy formula for a capacitor, we have:
E' = (1/2) * C * V^2
If the charge on each plate is doubled (Q' = 2Q), the energy stored can be calculated as:
E' = (1/2) * (2Q)^2/C
= (1/2) * 4Q^2/C
= 2 * (1/2) * Q^2/C
= 2E
So, in this case as well, we can see that the energy stored in the capacitor is doubled when the charge on each plate is doubled.
In summary:
(a) If the potential difference is doubled, the energy stored in the capacitor increases by a factor of 4.
(b) If the charge on each plate is doubled, the energy stored in the capacitor increases by a factor of 2.
To determine how the stored energy in a capacitor changes under different conditions, we need to understand the relationship between energy, potential difference (voltage), and charge. The energy stored in a capacitor is given by the equation:
E = (1/2) * C * V^2
Where E is the energy stored, C is the capacitance of the capacitor, and V is the potential difference across the capacitor.
(a) If the potential difference is doubled, we can calculate the change in energy by substituting the new value into the equation:
E' = (1/2) * C * (2V)^2
= (1/2) * C * 4V^2
= 2 * (1/2) * C * V^2
= 2E
Therefore, if the potential difference is doubled, the energy stored in the capacitor will also double. It means that the stored energy increases by a factor of 2.
(b) If the charge on each plate of the capacitor is doubled while the potential difference remains the same, we can use the formula for energy in terms of charge:
E = (1/2) * C * V^2
When the charge on each plate is doubled, the charge (Q) in the equation above is replaced with 2Q:
E' = (1/2) * C * V'^2
= (1/2) * C * (Q')^2
= (1/2) * C * (2Q)^2
= (1/2) * C * 4Q^2
= 2 * (1/2) * C * 2^2 * Q^2
= 4 * (1/2) * C * Q^2
= 4E
Hence, if the charge on each plate of the capacitor is doubled while the potential difference remains the same, the energy stored in the capacitor increases by a factor of 4.
In summary, when the potential difference across the capacitor is doubled, the energy stored in the capacitor also doubles (a factor of 2 increase). On the other hand, if the charge on each plate of the capacitor is doubled while the potential difference remains the same, the energy stored in the capacitor increases by a factor of 4.