For prectical applications , two or more capacitors are often used in combination and their total capacitance C must be known.To find total capacitance of the arrangement of capacitor we would use equation
Q=CV (i) Parallel combination of capacitors
Figure below shows two capacitors connected in parallel between two points A and B
Right hand side plate of capacitors would be at same common potential VA. Similarly left hand side plates of capacitors would also be at same common potential VB.
Thus in this case potential difference VAB=VA-VB would be same for both the capacitors, and charges Q1 and Q2 on both the capacitors are not necessarily equal. So,
Q1=C1V and Q2=C2V
Thus charge stored is divided amongst both the capacitors in direct proportion to their capacitance.
Total charge on both the capacitors is,
So system is equivalent to a single capacitor of capacitance
When capacitors are connected in parallel their resultant capacitance C is the sum of their individual capacitances.
The value of equivalent capacitance of system is greater then the greatest individual one.
If there are number of capacitors connected in parallel then their equivalent capacitance would be
C=C1+C2+ C3........... (10)
(ii) Series combination of capacitors
Figure 7 below shows two capacitors connected in series combination between points A and B.
Both the points A and B are maintained at constant potential difference VAB.
In series combination of capacitors right hand plate of first capacitor is connected to left hand plate of next capacitor and combination may be extended foe any number of capacitors.
In series combination of capacitors all the capacitors would have same charge.
Now potential difference across individual capacitors are given by
Sum of VAR and VRB would be equal to applied potential difference V so,
=Q(1/C1 + 1/C2)
i.e., resultant capacitance of series combination C=Q/V, is the ratio of charge to total potential difference across the two capacitors connected in series.
So, from equation 12 we say that to find resultant capacitance of capacitors connected in series, we need to add reciprocals of their individual capacitances and C is always less then the smallest individual capacitance.
Result in equation 12 can be summarized for any number of capacitors i.e.,
7. Energy stored in a capacitor
Consider a capacitor of capacitance C, completely uncharged in the begning.
Charhing process of capacitor requires expanditure of energy because while charging a capacitor charge is transferred from plate at lower potential to plate at higher potential.
Now if we start charging capacitor by transporting a charge dQ from negative plate ti the positive plate then work is done against the potential difference across the plate.
If q is the amount of charge on the capacitor at any stage of charging process and φ is the potential difference across the plates of capacitor then magnitude of potential difference is φ=q/C.
Now work dW required to transfer dq is
To charge the capacitor starting from the uncharged state to some final charge Q work required is
Integrating from 0 to Q
Which is the energy stored in the capacitor and can also be written as
From equation 14c,we see that the total work done is equal to the average potential V/2 during the charging process ,multiplied by the total charge transferred
If C is measured in Farads ,Q in coulumbs and V in volts the energy stored would in Joules
A parallel plate capacitor of area A and seperation d has capacitance
electric field in the space between the plates is
E=V/d or V=Ed
Putting above values of V and C in equation 14b we find
If u denotes the energy per unit volume or energy density then
u=(1/2)ε0E2 x volume
The result for above equation is generally valid even for electrostatic field that is not constant in space.
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