A) \[(10/9)\,Volt/\mu \,m\] Volt/ |li m and in the \[+\text{v}e\,x\]x direction
B) \[\left( 5/3 \right)\text{ }Volt/\mu \,m\]and in the \[-\text{v}ex\]direction
C) \[\left( 5/3 \right)\text{ }Volt/\mu \,m\]and in the \[+\text{v}e\text{ }x\]direction
D) \[\left( 10/9 \right)\text{ }Volt/\mu \text{ }m\]and in the \[-\text{v}e\text{ }x\]direction
Correct Answer: A
Solution :
Here, \[V(x)=\frac{20}{{{x}^{2}}-4}\]volt We know that \[E=-\frac{dV}{dx}=-\frac{d}{dx}\left( \frac{20}{{{x}^{2}}-4} \right)\] or \[E=+\frac{40x}{{{({{x}^{2}}-4)}^{2}}}\] At \[x=4\,\mu m,\] \[E=+\frac{40\times 4}{{{({{4}^{2}}-4)}^{2}}}=+\frac{160}{144}=+\frac{10}{9}Volt/\mu m\]. Positive sign indicates that \[\vec{E}\] is in \[+ve\]x-direction.You need to login to perform this action.
You will be redirected in
3 sec