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A10-cmlong thin glass rod uniformly charged to+10nCand a 10-cm-long thin plastic rod uniformly charged to-10nCare placed side by side, 4.0cmapart. What are the electric field strengthsE1toE3at distances1.0cm,2.0cm, and from the glass rod a3.0cmlong the line connecting the midpoints of the two rods?

Short Answer

Expert verified

Electric field at distance 3cmis22.84×104N/C.

Electric field at distance 2cmis 16.7×104N/C.

Electric field at distance1cmis22.84×104N/C.

Step by step solution

01

Figure for two rods with electric field strength

Figure is,

02

Calculation for Electric field strength at distance r=1 cm

The electric field produced by a rod at a distance perpendicular to its midpoint is calculated as follows:

E=kqrr2+l24

k=9×109N·m2/C2.

q-charge

The electric field at r=1cm:

An electric field is produced by both glass and plastic rods, and the direction of the field is determined by the charge on the rods. which is in the(+)xdirection.

We'll also include both electric fields because both rods are pointed in the same direction.

Electric field is,

E1=Eglass+Eplastic

=9×109N·m2/C210×10-9C(0.01m)(0.01m)2+0.12m2+9×109N·m2/C210×10-9C((0.04-0.01)m)((0.04-0.01)m)2+0.12m2

=(17.7+5.14)×104N/C

E1=22.84×104N/C

03

Calculation of electric field at distance r=2.0 cm and r=3.0 cm

Forr=2.0cm:

E2=Eglass+Eplastic

=9×109N·m2/C210×10-9C(0.02m)(0.02m)2+0.12m2+9×109N·m2/C210×10-9C((0.04-0.02)m)((0.04-0.02)m)2+0.12m2

=(8.35+8.35)×104N/C

E2=16.7×104N/C

For r=3.0cm:

E3=Eglass+Eplastic

=9×109N·m2/C210×10-9C(0.03m)(0.03m)2+0.12m2+9×109N·m2/C210×10-9C((0.04-0.03)m)((0.04-0.03)m)2+0.12m2

=(5.14+17.7)×104N/C

E3=22.84×104N/C

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