Answer
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Hint :Here, Capacitors are the components in the electric circuit which stores charge according to the capacity of the capacitor. The combinatorial formula for calculating the capacitance of capacitors connected in parallel is similar to that of the resistance equivalent we calculate in series. And vice versa. Use the proper circuit diagram according to the conditions given in the question.
Complete Step By Step Answer:
According to the condition given in the question let us draw a proper circuit diagram containing two capacitors in the parallel combination and a potential is applied to them as shown in the figure $ (1) $
First, let the charge on capacitor $ C $ be $ {q_1} = CV $
And charge on capacitor $ 2C $ be $ {q_2} = 2CV $
Also, it is given that the capacitor with capacitance $ C $ is filled with dielectric $ k $ and $ V' $ be the potential through both the capacitors such that as shown in the figure $ (2) $
Now, when dielectric is filled in capacitor $ C $ the charge becomes $ {q_1}' = kCV' $
And similarly for capacitor $ 2C $ charge becomes $ {q_2}' = 2CV' $
From figures $ (1) $ and $ (2) $ , we know that the total charge will remain conserved
That is why, we have to use the formula for conservation of charges in capacitors having same capacitance but different voltage across them to calculate new potential difference.
$ \therefore CV + 2CV = kCV' + 2CV' $
$ \therefore V' = \dfrac{{3V}}{{k + 2}} $
The new voltage across the capacitor $ C $ is $ V' = \dfrac{{3V}}{{k + 2}} $ .
Note :
We know that the definition says that the charges on a capacitor is the product of the capacity of the capacitor and the potential difference across it i.e. $ q = CV $ . We have used this formula to find the potential difference when the dielectric is added.
Complete Step By Step Answer:
According to the condition given in the question let us draw a proper circuit diagram containing two capacitors in the parallel combination and a potential is applied to them as shown in the figure $ (1) $
First, let the charge on capacitor $ C $ be $ {q_1} = CV $
And charge on capacitor $ 2C $ be $ {q_2} = 2CV $
Also, it is given that the capacitor with capacitance $ C $ is filled with dielectric $ k $ and $ V' $ be the potential through both the capacitors such that as shown in the figure $ (2) $
Now, when dielectric is filled in capacitor $ C $ the charge becomes $ {q_1}' = kCV' $
And similarly for capacitor $ 2C $ charge becomes $ {q_2}' = 2CV' $
From figures $ (1) $ and $ (2) $ , we know that the total charge will remain conserved
That is why, we have to use the formula for conservation of charges in capacitors having same capacitance but different voltage across them to calculate new potential difference.
$ \therefore CV + 2CV = kCV' + 2CV' $
$ \therefore V' = \dfrac{{3V}}{{k + 2}} $
The new voltage across the capacitor $ C $ is $ V' = \dfrac{{3V}}{{k + 2}} $ .
Note :
We know that the definition says that the charges on a capacitor is the product of the capacity of the capacitor and the potential difference across it i.e. $ q = CV $ . We have used this formula to find the potential difference when the dielectric is added.
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