Answer:
E" = Q/4πε₀√[(x² + R²)]³(x - (L - x)/√[(L - 2x)L/(x² + R²) + 1]³})
Explanation:
The electric field due to a charged ring of radius R at a distance x from the center of the ring when the axis of the ring is located on the x - axis is
E = Qx/4πε₀[√(x² + R²)]³
Since the rings are separated by a distance L, the electric field at point x due to the second ring is E' = -Q(L - x)/4πε₀[√((L - x)² + R²)]³. It is negative since it points in the negative x - direction.
So, the resultant electric field at x is E" = E + E' = Qx/4πε₀[√(x² + R²)]³ + {-Q(L - x)/4πε₀[√((L - x)² + R²)]³}
E" = Qx/4πε₀√[(x² + R²)]³ - Q(L - x)/4πε₀√[((L - x)² + R²)]³
E" = Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[((L - x)² + R²)]³})
E" = Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[(L² - 2Lx + x² + R²)]³})
E" = Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[(L - 2x)L + (x² + R²)]³})
E" = Q/4πε₀√[(x² + R²)]³(x - (L - x)/√[(L - 2x)L/(x² + R²) + 1]³})
So, the electric field at points along the x axis is
E" = Q/4πε₀√[(x² + R²)]³(x - {(L - x)/√[(L - 2x)L/(x² + R²) + 1]³})
