Question

Engineers discover that the electric potential between two electrodes can be modeled as $V\left(x\right)=V_{0}In(1+x/d)$, where $V_0$is a constant, $x$is the distance from the first electrode in the direction of the second, and $d$is the distance between the electrodes. What is the electric field strength midway between the electrodes?

Short answer

The electric field strength midway between the electrodes is$-\frac{2V_0}{3d}$.

Step-by-step solution

Given information 

We have given that the electric potential between two electrodes can be modeled as $V1x2=V0In11+x/d2,$where $V0$is a constant, $x$is the distance from the first electrode in the direction of the second, and d is the distance between the electrodes.

We need to find the the electric field strength midway between the electrodes.

Simplify 

The electric field strength as the negative of the gradient, which means in the one-dimensional case, the derivative.

$E\left(x\right)=-\frac{dV\left(x\right)}{dx}$

To find the field in a given position we just need to substitute this position in the expression. In other words, we need to now perform this derivation and using the result.

localid="1648621499660" $$\begin{gathered} E\left(x\right)=\frac{d}{dx}{V}_{0}In1+\frac{x}{d}=-V_0\frac{1}{1+{\displaystyle \frac{x}{d}}}\frac{d}{dx}1+\frac{x}{d} \\ E\left(x\right)=-V_0\frac{1}{{\displaystyle \frac{x+d}{d}}}\frac{1}{d} \end{gathered}$$,

Thus,
$\frac{E\left(x\right)=-{\displaystyle \frac{V_0}{x+d}}}{}$

As said before, we are just left to substitute $x=\frac{d}{2}$and find

localid="1648621455884" $$\begin{gathered} E\left(\frac{d}{2}\right)=-\frac{V_0}{{\displaystyle \frac{d}{2}}+d}=\frac{V_0}{{\displaystyle \frac{d+2d}{2}}} \\ =-\frac{2V_0}{3d} \end{gathered}$$