question_answer

Two infinitely long parallel conducting plates having surface charge densities + \[\Delta \phi =2\times 0.01\times \pi \times {{(1)}^{2}}\times 200\times \cos {{0}^{o}}\]and -\[\therefore \] respectively, are separated by a small distance. The medium between the plates is vacuum. It\[e=\frac{-2\times 0.01\times \pi \times {{(1)}^{2}}\times 200}{100}\] is the dielectric permittivity of vacuum, then the electric field in the region between the plates is:

A) zero                                      

B) \[=-4\pi Volt\]  

C)       \[\therefore \] 

D)       \[E=\frac{|e|}{2\pi r}=\frac{4\pi }{2\pi r}=\frac{2}{1}=2V/m\]

Correct Answer: C

Solution :

                Given that conducting plates have surface charge densities \[\therefore \]and \[I=\Delta p=m\frac{\Delta x}{\Delta t}\] respectively. Since the sheet is large, the electric field E at energy point near the sheet will be perpendicular to the sheet. The resultant electric field is given by \[m=0.1kg,\frac{\Delta x}{\Delta t}=\frac{4}{2}m/s\] If \[\therefore \] is surface charge density then, electric field \[=I=0.1\times \frac{4}{2}=0.2kg\,\,m{{s}^{-1}}\] \[=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{u}^{2}}={{K}_{f}}-{{K}_{i}}\]  \[\therefore \]