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 \[=I=0.1\times \frac{4}{2}=0.2kg\,\,m{{s}^{-1}}\] \[=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{u}^{2}}={{K}_{f}}-{{K}_{i}}\] \[\therefore \]You need to login to perform this action.
You will be redirected in
3 sec