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Suppose we have two parallel plates (each of area $A$) spaced with distance $d_1$ acting as a parallel-plate capacitor within a circuit. If the capacitor is connected to a power source with constant voltage $V_s$, positive charge will accumulate on the parallel plate connected to the positive terminal and negative charge will accumulate on the plate connected to the negative terminal. This will create an electrical potential between the two plates.

Does this process continue until the voltage $V_t$ between the plates is equal to $V_s$, and if not, how can we calculate the terminal voltage? Suppose we move the plates closer to distance $d_2$. Will the terminal voltage be identical to that of $d_1$?

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how can we calculate the terminal voltage?

As the problem is stated, the question answer itself:

the capacitor is connected to a power source with constant voltage $V_s$

(emphasis mine). By KVL, the voltage across the capacitor is the voltage across the power source which is the constant $V_s$.

Suppose we move the plates closer to distance d2. Will the terminal voltage be identical to that of d1?

Yes, since the voltage across the capacitor is the constant $V_s$.

However, since changing the distance between the plates changes the capacitance, there will be a current through the capacitor proportional to the rate of change of capacitance:

$Q = CV$

$\dfrac{dQ}{dt}= I_c = C\dfrac{dV_s}{dt} + V_s \dfrac{dC}{dt}$

Since the voltage is stipulated to be constant, the first term on the right hand side is zero.

However, even with a constant voltage across the capacitor, there can be a capacitor current due to time dependent capacitance.

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