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How  does  the  CHARGE, CAPACITY, POTENTIAL ,  ELECTRIC FIELD and  ENERGY  associated wih a  capacitor   changes  when  the  SEPERATION between  the PLATES  INCREASES [A] when BATTERY  IS  DISCONNECTED  [B] when  BATTERY  IS  CONNECTED
Asked by pardeepkumar2281 | 22 May, 2018, 10:35: PM
We use the following formulas for air filled parallel plate capacitor

(a)   ;
C - capacitance, ε0 is permitivity of free space, A - area of plates, d is separation of plates, V is potential difference, E = energy

First we charge the capacitor by connecting battery, to study the effect of separation distance on different quantities

(i) charge :- charge will decrease because capacitance is inversely proportional to separation distance and charge is proportional to capacitance
(ii) Capacity :- capacitance will decrease if we increase the separation distance
(iii) Potential  difference :- Potential depends on EMF of battery, nothing to do with separation distance
(iv) Electric field :- Electric field is potential divided by distance. Hence for a fixed potenial difference, field will decrease if separation distance increases
(v) Energy :- Energy is directly proportional to Capacitance for a given potential difference. Hence E will decrease if separation distance increases.

Now we disconnect the battery after fully charging the capacitor.

(i)  charge :- charge will not be affected
(ii) Capacitance will decrease
(iii) To keep the same charge and decreased capacitance, potential difference will decrease
(iv) Electric filed will decrease
(v) Energy will decrease

Answered by Thiyagarajan K | 06 Jul, 2018, 05:29: PM

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