Question

A total charge of Q coulomb is uniformly distributed along a rod 40cm in length.find find the electric field intensity 20cm away from the rod along its perpendicular bisector?

Answer 1

For a total charge of Q coulomb is uniformly distributed along a rod 40cm in length,  the electric field intensity 20cm away from the rod is mathematically given as

E1=1.598*10^11v/m

What is the electric field intensity 20cm away from the rod along its perpendicular bisector?

Generally, the equation for the  initial electric field intensity   is mathematically given as

$dEp=\int{kd/r}cosd\theta$

Therefore

$e1=kd/r{sin\theta2+sinR1}$

Hence

$E1=(9*10^9/20*10^{-2})({sin45+sin45})*B/40*10^-2$

E1=B*9*10^{13})/(10*110)*$\sqrt{2}$

E1=1.598*10^11v/m

In conclusion, the electric field intensity

E1=1.598*10^11v/m

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