Isn't E=changeinvoltages/meter ?
You can easily get the E in each direction, then add them as vectors.
A cube of side length L = 10 cm is centered at
the origin of an x,y,z (cartesian) coordinate system. The electric potential V (x, y, z) is measured
at the centers of the six faces of the cube to be the values shown below. Using this information,
estimate the magnitude and direction of the electric field at the center of
the cube.
V(L/2,0,0) = 10.0 Volts
V(-L/2,0,0) = -10.0 Volts
V(0,L/2,0) = 6.0 Volts
V(0,-L/2,0) = 9.0 Volts
V(0,0,L/2) = -4.0 Volts
V(0,0,-/2) = 6.0 Volts
2 answers
In the x direction, V increases by 20V in a distance L = 0.10 m, so the Ex field (which is - dV/dx) is -200 V/m.
Similarly, the y direction, Ey = 30 V/m, and in the z direction Ez = 100 V/m
The magnitude of E at the center is sqrt(200^2 + 30^2 + 100^2) = 225.6 V/m.
The field direction can be determined from the relative values of Ex, Ey and Ez. The ratios of Ex etc to |E| will provide direction cosines.
This is an estimate. The exact value will depend upon how potential is distributed over each surface, and may require "fringing" effects to be considered.
Similarly, the y direction, Ey = 30 V/m, and in the z direction Ez = 100 V/m
The magnitude of E at the center is sqrt(200^2 + 30^2 + 100^2) = 225.6 V/m.
The field direction can be determined from the relative values of Ex, Ey and Ez. The ratios of Ex etc to |E| will provide direction cosines.
This is an estimate. The exact value will depend upon how potential is distributed over each surface, and may require "fringing" effects to be considered.
1 answer