Answer:
The electric field at point P is [tex]4.27\times10^{5}\ N/C[/tex]
Explanation:
Given that,
Linear density [tex]\lambda_{1}=-7.8\ \mu C/m[/tex]
Linear density [tex]\lambda_{2}=5.4\ \mu C/m[/tex]
Inner radius a=2.9 cm
Outer radius = 5.3 cm
We need to calculate the electric field at point P
Using formula of electric field
[tex]E_{x}=\dfrac{(\lambda_{1}+\lambda_{2})}{2\pi\epsilon_{0}d}[/tex]
Put the value into the formula
[tex]E_{x}=\dfrac{(-7.8+5.4)\times10^{-6}}{2\pi\times8.85\times10^{-12}\times(-0.101)}[/tex]
[tex]E_{x}=4.27\times10^{5}\ N/C[/tex]
Hence, The electric field at point P is [tex]4.27\times10^{5}\ N/C[/tex]
