Calculating the Potential from the Field — Question 10

Question 10

A measured quadratic field fixes the cubic potential and its uncertainty sensitivity

On $0\le x\le L$, $E_x=E_0[1+\eta(x/L)^2]$. E0 and L are positive; $\eta$ is dimensionless. Set V(0)=0.

1. Find V(x) and the endpoint voltage drop D=V(0)-V(L).

2. Given $D=1.40E_0L$, infer $\eta$ and find the change in D from a small calibration error $\delta\eta$.

Question 10 - Solution

A measured quadratic field fixes the cubic potential and its uncertainty sensitivity

(a) Method and calculation

$$V(x)=-E_0\int_0^x[1+\eta(u/L)^2]du=-E_0\left(x+\frac{\eta x^3}{3L^2}\right).$$$$D=E_0L(1+\eta/3).$$

(b) Interpretation and further calculation

$$1+\eta/3=1.40\Rightarrow\eta=1.20.$$$$\delta D=\frac{E_0L}{3}\delta\eta,\qquad\frac{\delta D}{D}=\frac{\delta\eta}{4.20}.$$

Physical check

The endpoint voltage averages the field over the path; its sensitivity to the quadratic coefficient is reduced by the integral factor one third.