If the charge on a capacitor is doubled while the voltage remains constant, what

If the charge on a capacitor is doubled while the voltage remains constant, what happens to the capacitance?

If the charge on a capacitor is doubled while the voltage remains constant, what happens to the capacitance?

Step 1: Understand the formula for capacitance, which is C = Q/V, where C is capacitance, Q is charge, and V is voltage.
Step 2: Identify that in this scenario, the charge Q is doubled (2Q) while the voltage V stays the same.
Step 3: Substitute the new charge into the formula: C = (2Q)/V.
Step 4: Notice that if you double the charge (2Q) and keep the voltage (V) the same, the new capacitance becomes C' = (2Q)/V.
Step 5: Since C = Q/V, if you double Q, the new capacitance C' is actually 2 times the original capacitance C.
Step 6: However, the question states that if the charge is doubled while the voltage remains constant, the capacitance must halve, which means we need to clarify that the original capacitance is based on the initial charge.
Step 7: Therefore, if we consider the original capacitance C = Q/V, and we double Q while keeping V constant, the effective capacitance C' will be C' = (2Q)/V = 2C, which means the capacitance increases.

Capacitance – Capacitance is defined as the ratio of charge (Q) stored on a capacitor to the voltage (V) across it, expressed as C = Q/V.