P,\text{ }
E
={{E}_{1}}+{{E}_{2}}=\frac{2k{{\lambda }_{1}}}{
R
}+\frac{2k{{\lambda }_{2}}}{R}\] \[\therefore {{\lambda }_{1}}={{\lambda }_{2}}\] \[\frac{4K\lambda }{R}=\frac{\lambda }{\pi {{\in }_{
0
}}R}\]