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.You need to login to perform this action.
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