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CBSE Class 12-science Answered

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
answered-by-expert Expert Answer
We use the following formulas for air filled parallel plate capacitor
 
(a) begin mathsize 12px style C space equals space fraction numerator epsilon subscript 0 A over denominator d end fraction end style  ; begin mathsize 12px style q space space equals space C over V space space semicolon end style  begin mathsize 12px style E space equals space 1 half C V squared end style
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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