Answer
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Hint: You could recall some relation that relates the voltage with the capacitance and charge of a capacitor. Then you could see which among them remains constant for a given capacitor. Thereby, we will get a proportionality relation and hence we could see what happens on doubling the voltage.
Formula used:
Charge of capacitor,
$Q=CV$
Complete answer:
In the question, we are said that the voltage that is applied across a capacitor of capacitance C is doubled from V to 2V. This is followed by certain conditions and we are supposed to find the correct statement among them.
In order to answer this, let us recall the relation that relates the capacitance, voltage and charge of a capacitor. This relation is given by,
$Q=CV$ …………………………………… (1)
Here, Q is the charge of the capacitor, C is its capacitance and V is the voltage.
The fact we have to always keep in mind is that the capacitance of a particular capacitor has a constant value, that is, it remains the same on doubling the voltage.
Now from (1),
$Q\propto V$
Therefore, we could say the charge in a capacitor is doubled on doubling the voltage.
Hence, option C is found to be the answer.
Note:
Another significant quantity related to capacitor is its energy. The energy stored in a capacitor is given by,
$E=\dfrac{1}{2}C{{V}^{2}}$
Clearly, we see that the energy is directly proportional to the square of the voltage. So, on doubling the voltage the energy becomes four times the initial value.
Formula used:
Charge of capacitor,
$Q=CV$
Complete answer:
In the question, we are said that the voltage that is applied across a capacitor of capacitance C is doubled from V to 2V. This is followed by certain conditions and we are supposed to find the correct statement among them.
In order to answer this, let us recall the relation that relates the capacitance, voltage and charge of a capacitor. This relation is given by,
$Q=CV$ …………………………………… (1)
Here, Q is the charge of the capacitor, C is its capacitance and V is the voltage.
The fact we have to always keep in mind is that the capacitance of a particular capacitor has a constant value, that is, it remains the same on doubling the voltage.
Now from (1),
$Q\propto V$
Therefore, we could say the charge in a capacitor is doubled on doubling the voltage.
Hence, option C is found to be the answer.
Note:
Another significant quantity related to capacitor is its energy. The energy stored in a capacitor is given by,
$E=\dfrac{1}{2}C{{V}^{2}}$
Clearly, we see that the energy is directly proportional to the square of the voltage. So, on doubling the voltage the energy becomes four times the initial value.
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