For shown situation, in steady state condition ratio of charge stored in the first and last $n^{th}$ capacitor is
A$1 : (n+1)$
B$(n^2 +1) : (n^2 -1)$
C$(n+1) : 1$
D$1 : n$
Medium
Download our app for free and get started
D$1 : n$
d $\mathrm{Q}_{1}=\mathrm{EC}, \mathrm{Q}_{\mathrm{n}}=\mathrm{E}(\mathrm{nC})=\mathrm{nQ}_{1} \Rightarrow \frac{\mathrm{Q}_{1}}{\mathrm{Q}_{\mathrm{n}}}=\frac{1}{\mathrm{n}}$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
A capacitor $C$ is fully charged with voltage $V _{0}$ After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{ C }{2} .$ The energy loss in the process after the charge is distributed between the two capacitors is$.........$$CV _{0}^{2}$
A parallel plate capacitor has a parallel sheet of copper inserted between and parallel to the two plates, without touching the plates. The capacity of the capacitor after the introduction of the copper sheet is :
The area of each plate of a parallel plate capacitor is $20\,cm^2$ and separation between the plates is $2\,mm$. If dielectric strength of air is $3 \times 10^6\,V/m,$ the maximum possible value of emf of the battery, which can be connected across the plates of this capacitor and the corresponding charge on the plates is
In the following diagram the work done in moving a point charge from point $P$ to point $A$, $B$ and $C$ is respectively as $W_A$, $W_B$ and $W_C$ , then
A parallel plate capacitor having crosssectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of same area but thickness $d/2$ is inserted between the plates as shown in figure having dielectric constant $K (=4) .$ The ratio of new capacitance to its original capacitance will be,
The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2\,\mu \,F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu \,F$