Respuesta :
Answer:
[tex]E_1=3.3\times10^5\ \rm N/C\\E_2=1.9\times10^5\ \rm N/C\\E_3=2.4\times10^5\ \rm N/C\\[/tex]
Explanation:
Given:
- Length of the rod, L=10 cm
- Magnitude of the charge on the rods Q=11 nC
- The distance between the rods d=4.10 cm
We know that the Electric Field due to the rod at the distance r on the axis along the perpendicular bisector of the rod is given by
[tex]E=\dfrac{2k\lambda}{r}[/tex]
At r= 1 cm from the glass rod
So we have [tex]E_1=2k\lambda \left (\dfrac{1}{0.01}-\dfrac{1}{0.031} \right)\\\\E_1=2\times 9\times10^9 \dfrac{11\times10^{-9}}{0.1} \left (\dfrac{1}{0.01}-\dfrac{1}{0.031} \right)\\\\E_1=3.3\times10^5\ \rm N/C[/tex]
The electric Field will be in the direction of plastic rod from glass rod.
At r= 2 cm from the glass rod
So we have [tex]E_2=2k\lambda \left (\dfrac{1}{0.02}+\dfrac{1}{0.0321} \right)\\\\E_2=2\times 9\times10^9 \dfrac{11\times10^{-9}}{0.1} \left (\dfrac{1}{0.02}+\dfrac{1}{0.021} \right)\\\\E_2=1.9\times10^5\ \rm N/C[/tex]
The electric Field will be in the direction of plastic rod from glass rod.
At r= 3 cm from the glass rod
So we have [tex]E_3=2k\lambda \left (\dfrac{1}{0.03}+\dfrac{1}{0.011} \right)\\\\E_3=2\times 9\times10^9 \dfrac{11\times10^{-9}}{0.1} \left (\dfrac{1}{0.03}+\dfrac{1}{0.011} \right)\\\\E_1=2.4\times10^5\ \rm N/C[/tex]
The electric Field will be in the direction of plastic rod from glass rod.
