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Several important properties of perfect conductors in electrostatic situation are discussed.

Three equal charges are placed at the corners of an equilateral triangle.
Show that the electric field at the center is zero.

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qm-lec-20001

Two spheres, each of radius $R$ and carrying charge densities $+\rho$ and $-\rho$ respectively, are placed so that they partially overlap. The separation between the centers of the spheres is $D$. Show that
the field in the region of overlap is constant and find its value

                                            Mid Semester Examination∗

B.Sc. IInd                                                                                           Sem MM: 30

1. A circular disk of radius \(R\) carries a surface charge density \(\sigma=kr\). Find the potential at a point on the axis of the disk and distance \(d\) from the center of the disk.
2. Two grounded infinite conducting planes are kept along the \(XZ\) and \(YZ\) planes, see Fig.2. A charge \(Q\) is placed at (4,3) find the force acting on the charge \(Q\). 
3. Solve the boundary value problem in volume \(V\) bounded by semi-infnite planes (i) Plane 1:\(XZ\) plane extending to infinity in positive \(z\) and both positive and negative \(x\) directions.(ii)Plane 2: Another plane parallel to Plane 1 obtained by translating it to \(y=L\)(iii)Infinite strip: A strip lying in \(XY\) plane between \(0\le y \le L\). The boundary conditions required to be satisfied are \begin{equation} \phi(x,y,z)= \begin{cases} 0 & \text{for } y=0 \text{ and all } x, z\\ 0 &\text{for } y=L \text{ and all } x, z\\ 0 & \text{ as } z \to \infty \\ \cos(3\pi y/L) \sin(5\pi y/L) & \text{ for } z=0 \text{ and } 0\le y \le L \end{cases} \end{equation} 

An alpha particle travels in a circular path of radius $0.45$m in a magnetic field with $B=1.2$ w/m$^2$. Calculate (i) its speed (ii) its period of revolution, and (iii) its kinetic energy. Mass of proton particle = \(1.67\times 10^{-27}\)kg \(\approx 4\times M_p= 4\times938.27\) MeV.

Solution :

• [(i)] the magnetic force \(eBv\) must be equal to the mass times acceleration. Therefore \begin{equation*} Bev = \frac{Mv^2}{R}, \end{equation*} where \(R\) is the radius of the circular orbit. Hence \begin{equation*} v= \frac{eBR}{M} = \frac{2\times1.6 \times10^{-19}\times 1.2 \times 0.45}{4\times 1.67\times 10^{-27}}\approx 2.7 \times10^7 \text{m/s}. \end{equation*}
• [(ii)] The time period is \begin{equation*} T = \frac{2\pi R}{v} = \frac{2\times3.14\times 0.45}{2.7\times10^7} \approx10^{-7} \text{ s}. \end{equation*}
• [(iii)] The kinetic energy is given by \begin{eqnarray}\nonumber \text{K.E.} &=& \frac{1}{2} M v^2= \frac{1}{2}\times (4\times 1.67 \times 10^{-27}) \times \big(2.7\times10^7\big)^2 \\\nonumber &=& 3.26\times 7.29 \times 10^{-13} \approx 23.7 \times 10^{-13} \text{J}. \end{eqnarray}