A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference between the plates and has charge Q1. If the separation between the plates is doubled, what will the magnitude of the charge Q2 be in relationship to Q1.

doubled

To determine the relationship between the magnitude of the charge Q2 and Q1 when the separation between the plates of the capacitor is doubled, we need to consider the capacitance of the capacitor.

The capacitance of a parallel plate capacitor is given by the formula:

C = ε₀ * (A / d)

where C is the capacitance, ε₀ is the permittivity of free space (a constant value), A is the area of the plates, and d is the separation between the plates.

When the separation between the plates is doubled, the new separation (d') becomes 2d.

The new capacitance (C') can be calculated by substituting the new separation into the formula:

C' = ε₀ * (A / 2d)

Now, we can determine the relationship between Q1 and Q2 by using the formula:

C = Q / V

where Q is the charge stored in the capacitor and V is the potential difference across the plates.

Since the battery maintains a constant potential difference between the plates, V is also constant for both situations.

Rearranging the formula, we have:

Q = C * V

Now, we can compare Q1 and Q2 by substituting the respective capacitance and the constant potential difference values for both cases:

Q1 = C * V
Q2 = C' * V

Substituting the capacitance values we derived earlier:

Q1 = (ε₀ * (A / d)) * V
Q2 = (ε₀ * (A / 2d)) * V

From these equations, we can see that the magnitude of the charge Q2 is half of the magnitude of Q1. So, when the separation between the plates of the capacitor is doubled, the magnitude of the charge Q2 will be half of the magnitude of Q1.