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A parallel plate capacitor is charged and then isolated. The effect of increasing the plate
separation on charge, potential and capacitance respectively are
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separation on charge, potential and capacitance respectively are
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Verified Answer
The correct answer is:
constant, increases, decreases
Given, parallel plate is charged and then isolated. Therefore, \( Q= \) constant.
Now, capacitance is given as \( C=\frac{\varepsilon_{0} A}{d} \Rightarrow C \propto \frac{1}{d} \)
So, if the separation between plates is increased then, C decreases.
Also, we know \( Q=C V \Rightarrow V=\frac{Q}{C} \Rightarrow V \propto \frac{1}{C} \)
So, if \( C \) decreases then potential increases.
Thus, on increasing the plate separation charge remains constant, potential increases and capacitance decreases.
Now, capacitance is given as \( C=\frac{\varepsilon_{0} A}{d} \Rightarrow C \propto \frac{1}{d} \)
So, if the separation between plates is increased then, C decreases.
Also, we know \( Q=C V \Rightarrow V=\frac{Q}{C} \Rightarrow V \propto \frac{1}{C} \)
So, if \( C \) decreases then potential increases.
Thus, on increasing the plate separation charge remains constant, potential increases and capacitance decreases.
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