Physics 2 Electricity and Magnetism

Electric potential energy is the energy stored in an electric field due to the positions of charged objects. When dealing with a parallel plate capacitor, which consists of two plates separated by a distance and charged with equal and opposite charges, the electric potential energy can be calculated using the properties of the electric field and the charges on the plates. Let's assume we have a parallel plate capacitor with plate area \( A \), separation distance \( d \), charge \( Q \) on one plate and \( -Q \) on the other plate, and an electric field \( E \) between the plates. The electric field \( E \) in a parallel plate capacitor is uniform and given by: \[ E = \frac{V}{d} \] where \( V \) is the potential difference between the plates. The potential difference \( V \) can also be related to the charge \( Q \) and the capacitance \( C \) of the capacitor by the equation: \[ V = \frac{Q}{C} \] The capacitance \( C \) of a parallel plate capacitor is given by: \[ C = \frac{\varepsilon_0 A}{d} \] where \( \varepsilon_0 \) is the permittivity of free space. Combining the above equations, we can express the electric field \( E \) in terms of the charge \( Q \) and the capacitance \( C \): \[ E = \frac{Q}{C d} = \frac{Q}{\varepsilon_0 A} \] Now, the electric potential energy \( U \) stored in a capacitor is given by: \[ U = \frac{1}{2} Q V \] Substituting \( V \) with \( Q/C \), we get: \[ U = \frac{1}{2} \frac{Q^2}{C} \] And substituting \( C \) with \( \varepsilon_0 A/d \), we obtain the expression for electric potential energy in terms of charge, distance, and area: \[ U = \frac{1}{2} \frac{Q^2 d}{\varepsilon_0 A} \] If the distance \( d \) between the plates is changed, the capacitance \( C \) will change accordingly, which in turn will affect the electric potential energy \( U \). Specifically, if the distance \( d \) is increased, the capacitance \( C \) decreases, and the electric potential energy \( U \) increases, assuming the charge \( Q \) remains constant. Conversely, if the distance \( d \) is decreased, the capacitance \( C \) increases, and the electric potential energy \( U \) decreases. In summary, the electric potential energy of a parallel plate capacitor is directly proportional to the square of the charge on the plates and the distance between the plates, and inversely proportional to the permittivity of free space and the area of the plates. When the distance between the plates changes, the electric potential energy changes as well, following the relationship described by the equation above.

Physics

How does the electric potential energy change when two plates separated by charge are moved to a distance d, and what is the mathematical expression for this energy in terms of charge, distance, and electric field?

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How does the electric potential energy change when two plates separated by charge are moved to a dis...

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