Equipotential Surfaces and Application of Electric Potential

The Equipotential Surface�DEFINED BY

It takes NO work to move a charged particle

between two points at the same potential.

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The locus of all possible points that require NO

WORK to move the charge to is actually a surface.

Example: A Set of Equipotenital Surfaces

Field Lines and Equipotentials

Equipotential

Surface

Electric

Field

Components

Equipotential

Surface

Electric

Field

E_normal

E_parallel

Δx

Work to move a charge a distance

Δx along the equipotential surface

Is Q x E_parallel X Δx

BUT

• This an EQUIPOTENTIAL Surface
• No work is needed since ΔV=0 for such a surface.
• Consequently E_parallel=0
• E must be perpendicular to the equipotential surface

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Therefore

E

E

E

V=constant

Field Lines are Perpendicular to the Equipotential Lines

Equipotential

Consider Two Equipotential�Surfaces – Close together

a

b

Work to move a charge q from a to b:

E

Where

Keep in Mind

• Force and Displacement are VECTORS!
• Potential is a SCALAR.

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UNITS

• 1 VOLT = 1 Joule/Coulomb
• For the electric field, the units of N/C can be converted to:
• 1 (N/C) = 1 (N/C) x 1(V/(J/C)) x 1J/(1 NM)
• Or

1 N/C = 1 V/m

• So an acceptable unit for the electric field is now Volts/meter. N/C is still correct as well.

In Atomic Physics

• It is sometimes useful to define an energy in eV or electron volts.
• One eV is the additional energy that an proton charge would get if it were accelerated through a potential difference of one volt.
• 1 eV = e x 1V = (1.6 x 10^-19C) x 1(J/C) = 1.6 x 10^-19 Joules.
• Nothing mysterious.

Potential due to Point Charge

Consider a unit charge (+) being brought from infinity to a distance r from a

Charge q:

q

r

To move a unit test charge from infinity to the point at a distance r from the charge q, the external force must do an amount of work that we now can calculate.

x

The math….

This thing must

be positive anyway.

r_final<r_initial🡪neg sign

Potential due to number of charges

Electric potential due to finite line of charge

P

At P

Using table of integrals

What about a rod

that goes from –L to +L??

Which was the result we obtained earlier

disk

σ=charge

per

unit

area

• THANKS