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A straight, nonconducting plastic wire 8.00 cm long carries a charge density of 100 nC/m distributed uniformly along its length. It is lying on a horizontal tabletop.
(a) Find the magnitude and direction of the Electric Field that is produced 6.0cm directly above its midpoint.(b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the Electric field on a point 6.0cm directly above its center.

Respuesta :

Answer:

(A) 16641 N/C (B) 72000 N/C

Explanation:

We have given the length of the wire = 8 cm =0.08 m

Charge density = 100 nC/m

So charge = 100 ×0.08 = 8 nC [tex]=8\times 10^{-9}C[/tex]

(A) Electric field E that produce 6 cm directly above its midpoint will be

[tex]E=\frac{KQ}{X\sqrt{X^2+a^2}}=\frac{9\times 10^9\times 8\times 10^{-9}}{0.06\sqrt{0.06^2+(\frac{0.08}{2})^2}}=16641N/C[/tex]

Here X is the distance where we have to find the electric field

(B) Now electric field due to flat ring will be

[tex]E=\frac{KQX}{(X^2+a^2)^\frac{3}{2}}[/tex]

Here X is the distance where we have to find the electric field

So [tex]E=\frac{9\times 10^9\times 8\times 10^{-9}}{(0.06^2+0.08^2)^\frac{3}{2}}=72000N/C[/tex]

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