Question

Four condensers having capacities $2pF$, $3pF$,$4pF$and $6pF$are connected in series. The equivalent capacitance of the combination is
(A) $8pF$
(B) $0.8pF$
(C) $1.8pF$
(D) $0.4pF$

Answer 1

Hint: We will be using the concept that when the same potential difference is applied across capacitors which are connected in series, then each capacitor has the same charge and the applied potential is equal to the sum of potential differences on each capacitor.
Formula Used: $V = Q/C$

 Complete step by step answer
let us say potential difference across $2pF$be ${V_1}$
potential difference across $3pF$ be ${V_2}$
potential difference across $4pF$ be ${V_3}$
potential difference across $6pF$ be ${V_4}$
we know that $V = \dfrac{q}{C}$, where q is the charge and C is the capacitance
total potential V is sum of all these four potentials $V = {V_1} + {V_2} + {V_3} + {V_4}$
\[ \Rightarrow V = q\left( {\dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}} + \dfrac{1}{{{C_4}}}} \right)\]
\[ \Rightarrow \dfrac{q}{V} = \left( {\dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}} + \dfrac{1}{{{C_4}}}} \right) \Rightarrow 1/{C_{eq}} = \left( {\dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}} + \dfrac{1}{{{C_4}}}} \right)\]
On substituting the values of different capacitance,
\[\dfrac{1}{{{C_{eq}}}} = \left( {\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{6}} \right)\]
\[ \Rightarrow {C_{eq}} = 0.8pF\]

 The equivalent capacitance of the combination is \[0.8pF\]
Correct answer is B. \[0.8pF\]

 Additional information
Capacitor is nothing but a pair of two conductors which can be of any shape are close to each other and have opposite charges.Capacitance is the ratio of charge on a capacitor plate to the potential difference between the plates. Capacitance and charge are proportional that is more the charge, greater is the capacitance.

 Note
There are few points to consider about capacitors in series like we need to observe the charge shifting from one to other capacitor in a series combination, it can move in one direction only, if not then there is no series combination. Another point is the battery attached in the circuit can produce charge on only that capacitor which is connected directly with the battery. Charges on other capacitors are due to shifting of present charge. The charge can only be redistributed.