Physics Electrostatic Potential and Capacitance

Ten charges are placed on the circumference of a circle of radius $10\mathrm{s}$ $$ with constant angular separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q)$$ each, while $\mathrm{2,4},\mathrm{6,8},10$ $$ have charge $$ (-q)each. The potential v$$ and the electric field $$ E at the centre of the circle are respectively: (Take V = 0 $$ at infinity)

In the circuit shown, charge on the $5\mu \mathrm{F}$ $$ capacitor is:

Three charged particles $A,B$ $$ and $C$ $$ with charges $-4q,2q$ $$ and $$ $-2q$  are present on the circumference of a circle of radius $d$ $$ . The charged particles $A,C$ $$ and centre $O$ $$ of the circle formed an equilateral triangle as shown in figure. Electric field at $\mathrm{O}$ $$ along $\mathrm{x}$ $$ -direction is:

A $60\mathrm{p}\mathrm{F}$ capacitor is fully charged by a $20\mathrm{V}$ supply. It is then disconnected from the supply and is connected to another uncharged $60\mathrm{p}\mathrm{F}$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $\mathrm{n}\mathrm{J}$  )

In the figure, potential difference between $A$ and $B$  is:

A parallel plate capacitor has plates of area A separated by distance ' $d$ $$ ' between them. It is filled with a dielectric which has a dielectric constant that varies as $k\left(n\right)=k(1+$ $\alpha n)$ $$ $$ where ' $$ $x$ ' is he distance measured from one of the plates.

If $\left(\alpha \mathrm{d}\right)<<1$ $$ , the total capacitance of the system is best given by the expression:

In a parallel plate capacitor set up, plate area of capacitor is 2 m² and the plates are separated by 1m. If the space between the plates are filled with a dielectric material of thickness 0.5 m and area 2 m² (See fig) the capacitance of the set-up will be ______ ε₀. (Dielectric constant of the material = 3.2) (Round off to the Nearest Integer)

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectricconstant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is 3d/4 , where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation:

Two capacitors of capacities 2C and C are joined in parallel and charge up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be:

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be: (Given area of plate = A)