Answer:
The magnitude of the electric field is 8504 N/C
Explanation:
rc = 0.35 m, Ac = 2πr^2 = 2 x π x 0.35^2 = 0.769 m^2, σ = 1.6 μC/m2 = 0.0000016 μC/m2
Surface Charge Density = total charge / surface area cylinder
σ = q / A ⇒ q = σ · A
q = 0.0000016 x 0.769
q = 1.23 μC
r = 1.3 m, k = 8.988 x 10^9
electric field vector = Coulomb constant * charge ÷ distance from charge
E = k * q ÷ r
E = 8.988 x 10^9 x 1.23 x 10^(-6) ÷ 1.3
E = 8504 N/C
Given values:
Now,
The Surface Area of cylinder be:
= [tex]2 \pi r^2[/tex]
By putting the values,
= [tex]2\times \pi\times (0.35)^2[/tex]
= [tex]0.769 \ m^2[/tex]
The Surface Charge Density (σ),
= [tex]\frac{Total \ charge}{Surface \ area \ cylinder}[/tex]
or,
⇒ [tex]q = \sigma.A[/tex]
[tex]=0.0000016\times 0.769[/tex]
[tex]= 1.23 \ \mu C[/tex]
hence,
The Magnitude will be:
⇒ [tex]Electric \ field \ vector = \frac{Coulomb \ constant}{Distance}[/tex]
or,
⇒ [tex]E = \frac{k\times q}{r}[/tex]
By substituting the values,
[tex]= \frac{8.988\times 10^9\times 1.23\times 10^{-6}}{1.3}[/tex]
[tex]= 8504 \ N/C[/tex]
Thus the approach above is right.
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