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Q) A straight, nonconducting plastic wire 8.50 cm long carries a charge density of +175 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 6.00 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 6.00 cm directly above its center.

Short Answer

Expert verified

Answer:

(a) The magnitude of the electric field at point a point 6.00cm directly above its midpoint is3.04×104N/Cand the direction is in positive x direction.

(b) The magnitude of the electric field at point a point 6.00cm directly above its center is 3.46×10-4N/Cand the direction of electric field is in +x direction .

Step by step solution

01

Formula related to the electric field of the ring

Electric Field of a Line Segment at distance x above the midpoint of a straight line segment:

E=kQx(x2+a2)

where k=14πε=9.0×109N.m2/C2, Q=charge on the ring and a= radius of the ring

02

Calculating the radius of the ring and charge on the ring

The length a is half the total length

a=L2=8.52×10-2=4.25×10-2m/s

The total charge is

Q=λL=175×10-9×8.5×10-2=1.49×10-8C

Substitute the values to find E:

E=9×109×1.49×10-80.06(0.06)2+(0.425)2=3.04×104N/C

Since the charge is positive so the field line is in positive x direction

03

If the wire is forming a ring

If the wire is formed to a loop then the electric field at a distance x on the axis of a uniformly charged ring is:

E=kQx(x2+a2)3/2

where k=14πε=9.0×109N.m2/C2, Q=charge on the ring and a= radius of the ring

The radius a is given by

2πa=La=8.5×10-22π=1.35×10-2m

Substitute the values to find E:

E=9×109×1.49×10-9×6×10-2(0.062+0.01352)3/2=3.46×10-4N/C

Since the charge is positive so the field line is in +x direction.

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