Calculate the area of the plates of a one farad parallel plate capacitor if the separation between plates is $1 \mathrm{mm}$ and plates are in a vacuum.

1.  $18\times {10}^8 {\mathrm{m}}^2$                 

2.  $0.3\times {10}^8 {\mathrm{m}}^2$

3.  $1.3\times {10}^8 {\mathrm{m}}^2$               

4.  $1.13\times {10}^8 {\mathrm{m}}^2$

A solid conducting sphere of radius a having a charge $\mathrm{q}$ is surrounded by a concentric conducting spherical shell of inner radius $2\mathrm{a}$ and outer radius $3\mathrm{a}$ as shown in figure.

Find the amount of heat produced when switch is closed.

1. ${\mathrm{Kq}}^2/2\mathrm{a}$                         

2. ${\mathrm{Kq}}^2/3\mathrm{a}$

3. ${\mathrm{Kq}}^2/4\mathrm{a}$                         

4. ${\mathrm{Kq}}^2/6\mathrm{a}$

A parallel plate capacitor of area ‘A’ , plate separation ‘d’ is filled with two dielectrics as shown. What is the capacitance of the arrangement ?

1.  $\frac{3{\mathrm{K\epsilon }}_0\mathrm{A}}{4\mathrm{d}}$                             

2. $\frac{4{\mathrm{K\epsilon }}_0\mathrm{A}}{3\mathrm{d}}$

3. $\frac{\left(\mathrm{K}+1\right){\mathrm{\epsilon }}_0\mathrm{A}}{2\mathrm{d}}$                       

4.$\frac{\mathrm{K}\left(\mathrm{K}+3\right){\mathrm{\epsilon }}_0\mathrm{A}}{2\left(\mathrm{K}+1\right)\mathrm{d}}$

Assertion : A particle $\mathrm{A}$ of mass m and charge $\mathrm{Q}$ moves directly towards a fixed particle $\mathrm{B}$, which

                  has charge $\mathrm{Q}$. The speed of $\mathrm{A}$ is $\mathrm{v}$ when it is far away from $\mathrm{B}$. The minimum separation

                  between the particles is proportional to ${\mathrm{Q}}^2$.

Reason : Total energy remains conserved.

Assertion : Potential difference between two points lying in a uniform electric field may be equal

                  to zero.

Reason : Points of line normal to electric field is equipotential line. 

Assertion :  At a distance r from a +ve charged particle potential is V and electric field strength is E.

                   Then graph between ${\mathrm{V}}^2$ & $\mathrm{E}$ is straight line.

Reason : For a point charge $\frac{{\mathrm{V}}^2}{\mathrm{E}}=\frac{\mathrm{q}}{4\mathrm{\pi }{\in }_0}$

Assertion : Electrons move away from a region of lower potential to a region of higher potential.

Reason : Since an electron has a negative charge. 

Assertion :  Each of the plates of a parallel-plate capacitor is given equal positive charge Q. The

                   charges on the facing surfaces will be same. 

Reason : A negative charge (–Q) will be induced on each of the facing surfaces.

Assertion : Electric potential is a property of an electric field, regardless of whether a charged

                  object has been placed in that field.

Reason : Potential depends on test charge.