The Equipotential Surfaces Solved Equation, Exercises of Electrical Engineering

This is solution to problems related Electrical Engineering course. It was suggested by Prof. Bhooshan Sawhney at Shree Ram Swarup College of Engineering and Management. It includes: Magnitude, Potential, Energy, Change, Electric, Charges, Magnetic, Magnitude

Typology: Exercises

2011/2012

Uploaded on 07/20/2012

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5. The electric field produced by an infinite sheet of charge has magnitude E=σ/2ε0,whereσis the
surface charge density. The field is normal to the sheet and is uniform. Place the origin of a coordinate
system at the sheet and take the xaxis to be parallel to the field and positive in the direction of the
field. Then the electric potential is
V=Vsx
0
Edx=VsEx ,
where Vsis the potential at the sheet. The equipotential surfaces are surfaces of constant x; that is,
they are planes that are parallel to the plane of charge. If two surfaces are separated by xthen their
potentials differ in magnitude by V=Ex=(σ/2ε0)∆x.Thus,
x=2ε0V
σ=28.85 ×1012 C2/N·m2(50 V)
0.10 ×106C/m2=8.8×103m.
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  1. The electric field produced by an infinite sheet of charge has magnitude E = σ/ 2 ε 0 , where σ is the surface charge density. The field is normal to the sheet and is uniform. Place the origin of a coordinate system at the sheet and take the x axis to be parallel to the field and positive in the direction of the field. Then the electric potential is

V = Vs −

∫ (^) x

0

E dx = Vs − Ex ,

where Vs is the potential at the sheet. The equipotential surfaces are surfaces of constant x; that is, they are planes that are parallel to the plane of charge. If two surfaces are separated by ∆x then their potentials differ in magnitude by ∆V = E∆x = (σ/ 2 ε 0 )∆x. Thus,

∆x =

2 ε 0 ∆V σ

  1. 85 × 10 −^12 C^2 /N · m^2

(50 V)

  1. 10 × 10 −^6 C/m^2

= 8. 8 × 10 −^3 m.

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