Vidyamandir Classes Capacitors (b) Parallel Combinations : When capacitors are connected in parallel, the potential difference V across each is same and the charge on C 1 , C 2 is different i.e., Q 1 and Q 2 . E quivalent capacitance between a and b is : C = C 1 + C 2 + C 4 + . . . . . The charges on capacitors is given as : 1 1 1 2 C Q Q C C and 2 2 1 2 C Q Q C C 4. A parallel plate capacitor contains two dielectric slabs of thickness d 1 , d 2 and dielectric constants k 1 and k 2 respectively. The area of the capacitor plates and slabs is equal to A . Considering the capacitor as a combination of two capacitors in series, the equivalent capacitance C is given by : 0 1 2 1 2 A C d d k k In general for more than one dielectric slab : C = 0 i i A d k If V is the potential difference across the plates, the electric fields in the dielectrics are given as : 1 1 2 1 1 2 1 V E d d k k k 2 1 2 2 1 2 1 V E d d k k k 5.(a) If there exits a dielectric slab of thickness t inside a capacitor whose plates are separated by distance d , the equivalent capacitance is given as : 0 A C t d t k The equivalent capacitance is not affected by changing the distance of slab from the parallel plates. If the slab is of metal, the equivalent capacitance is : 0 A C d t (for a metal, k = ) + d 1 d 2 K 1 K 2 + + + + + + – – – – – – – Things to Remember Self Study Course for IITJEE with Online Support 25
Vidyamandir Classes Capacitors (b) Consider a capacitor with two dielectric slabs of same thickness d placed inside it as shown. The slabs have dielectric constants k 1 and k 2 and areas A 1 and A 2 respectively. Treating the combination as two capacitors in parallel, 0 1 1 2 2 C k A k A d 6. Consider a circuit where an uncharged capacitor C is connected to a cell of emf E through a resistance R and a switch S as shown. At t = 0, the switch S is closed. Let q , V c be the charge and voltage on the capacitor at time t and i be the current. This equation gives the expression for charge on capacitor as a function of time. The charge grows on the plate exponentially as shown on the graph. Note the following points. 1. In steady state : t and q C E 2. The voltage across capacitor also grows exponentially towards E . (1 ) t / RC q V E e C 3. The time constant ( ) of the circuit is defined as the time after which the charge has grown upto (1 – 1/ e ) = 0.63 63 % of its steady-state value. = R C 7. If we connect a charged capacitor C across a resistance R , the capacitor begins to discharge through R . The excess positive charge on high potential plate flows through R to the negative plate and in steady state, the capacitor plates become uncharged.
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