Interactive 2
The electrostatic field is conservative, or, curl-free: . Since the curl of any gradient is identically zero, its E-field vector can be expressed as the gradient of a scalar. This scalar is the electrostatic potential V:
, V/m.
Inversely,
, V,
where P is the observation point and is a reference point (where V is assumed equal to zero).
In Figure 1, an equipotential plot is given of the electrostatic potential distribution in the cross-section of a twin-lead line whose left wire is at 100 V potential and its right wire is at –100 V. The blue-green line along the x-axis is a streamline since it is orthogonal to every equipotential line it crosses. It is along —the direction of the fastest rate of change of the potential V for each point belonging to the line. We can then find E along this line as

where is the directional derivative of V with respect to x.
We can approximate the derivative by very small (but not infinitesimal) increments:

where is a very short line segment along x, and is the corresponding change in V. The shorter , the better the above approximation. Thus, if ( and being the begin and end points of the segment) then
.
The above derivative estimate is the most accurate at the point , which is right in the middle of the segment.
In our case, the equipotentials are plotted with 10-V increments, i.e. V. Note that the potential distribution is anti-symmetrical with respect to the y-axis. Figure 1 also displays the rectangular grid: the major grid points are spaced at 1 mm, while the minor grid points are every 0.2 mm. Thus, it allows accurate measurement of the length of the segment along the blue-green line.
Follow the instructions below in order to find the magnitude of the electric field at a point along the blue-green line, and compare your result with a highly accurate numerical calculation. |