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A straight, nonconducting plastic wire 9.50cm {\rm cm} long carries a charge density of 125nC/m {\rm nC/m} distributed uniformly along its length. It is lying on a horizontal tabletop.
A) Find the magnitude and direction of the electric field this wire produces at a point 5.50 {\rm cm} 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 it produces at a point 5.50cm {\rm cm} directly above its center.

Answer :

Bharathparasad577go

Part A. The midway magnitude of the electric field is 17.2 * 10³ N/C.

Part B. The electric field intensity is 47.17 * 10³ N/C N/C when the wire shape is a circle.

The intensity of an Electric Field

Given this, the wire length is 9.50 cm and the charge density is 125 nC/m.

Part A

The electric field at the wires halfway is

E = q/4πε * 1/ (z(z²/a²+1)¹/²)

Where E is the intensity of the electric field, 8.85*10⁻¹² is the permittivity, z is 5.5 cm, a is the midpoint, 9.5/2 cm, and q is the charge density.

Substituting the value in the above equation

E = 17.2 * 10³N/C

The midway magnitude of the electric field is 17.2 * 10³ N/C

Part B

When the wire is a circle then, the charge density at the wire is

Q = q*l

Q= 1.235 * 10⁻⁸ * 0.095

Q = 1.235*10⁻⁸ C

And to calculate the radius

r = l/2π

r = 9.50/ 2*3.14

r = 1.512cm

r = 0.015m

The electric field density at the midpoint of the circle is,

E = Q/4π * z(z²+r²)¹/² * E

Substituting the values in the above equation, we get

E = 47.17 * 10³ N/C

The electric field intensity is 47.17 * 10³ N/C N/C when the wire shape is a circle.

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