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Sagot :
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.
Learn more about electric field intensity here:
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