When capacitor 2 is removed from a parallel circuit, what is the effect on the voltage drop across capacitor 1 and the charge stored on it?

In a parallel circuit, all capacitors share the same voltage. When one capacitor is removed, such as capacitor 2 in this scenario, the voltage across the remaining capacitors remains unchanged. Since capacitor 1 is still connected in parallel, it maintains the same voltage drop as before.

The charge stored on a capacitor is defined by the formula ( Q = C \times V ), where ( Q ) is the charge, ( C ) is the capacitance, and ( V ) is the voltage across the capacitor. Since the voltage across capacitor 1 does not change when capacitor 2 is removed, the charge stored on capacitor 1 also remains constant, assuming that the capacitance of capacitor 1 does not change.

Therefore, removing capacitor 2 from the parallel setup does not affect the voltage drop across capacitor 1, nor does it alter the amount of charge stored on it.