Question

The potential at a point x (measured in $\mu m)$ due to some charges situated on the x -axis is given by $\frac{V\left(x\right)=20}{(x^2-4)}$ Volts. The electric field E at $x=4\mu m$ is given by

• A. $\frac{5}{3}Volt/\mu andinthe-vexdirection$
  
• B. $\frac{5}{3}Volt/\mu andinthe+vexdirection$
  
• C. $\frac{10}{9}Volt/\mu andinthe-vexdirection$
  
• D. $\frac{10}{9}Volt/\mu andinthe+vexdirection$

Answer 1

The correct option is D $\frac{10}{9}Volt/\mu andinthe+vexdirection$

$v\left(x\right)=\frac{20}{x^2-4}volts{E}_x=-\frac{\partial v}{\partial x}=\frac{20.-2x}{(x^2-4{)}^2}=\frac{40x}{(x^2-4{)}^2}$
At x =4 cm, $E_x=\frac{40\times 4}{{12}^2}=\frac{160}{12\times 12}=\frac{10}{9}$