Q141 / 180 · KCET 2015 · Physics

A parallel plate capacitor is charged and then isolated. The effect of increasing the plate
separation on charge, potential and capacitance respectively are

  • A
    constant, decreases, decreases
  • B
    increases, decreases, decreases
  • C
    constant, decreases, increases
  • Correct answer
    D
    constant, increases, decreases

Solution

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.

Hence, the correct answer is option: 'D'.

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