If the charge on a capacitor is doubled while the voltage remains constant, what happens to the capacitance?
Practice Questions
1 question
Q1
If the charge on a capacitor is doubled while the voltage remains constant, what happens to the capacitance?
It doubles
It halves
It remains the same
It quadruples
Capacitance C = Q/V; if Q is doubled and V remains constant, C must halve.
Questions & Step-by-step Solutions
1 item
Q
Q: If the charge on a capacitor is doubled while the voltage remains constant, what happens to the capacitance?
Solution: Capacitance C = Q/V; if Q is doubled and V remains constant, C must halve.
Steps: 7
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.