Download Electric Potential and Energy

Electric Potential
and Energy
Electric Potential energy
Recall that work is done to change an object’s energy type or energy
value.
 In electrical circuits, batteries do work to give charges electric
potential energy which they can use to do work on loads like lamps or
resistors, etc.
 The movement of charge through an electric field (parallel to the
field lines) means the field will do work to change the charge’s
electric potential energy.
 When a charge moves perpendicular to the field lines NO work is
done.

Electric Potential energy

Work is the dot product of force and displacement. Thus:
Ee = Fe●r
= kq1q2 ●r
r2
= kq1q2
r
 The
electric potential energy approaches zero as r
approaches infinity.
 An
electric potential energy between opposite charges
is negative; this is not a direction. (Energy is a scalar).
Electric Potential
 Electric
potential (or Voltage) is a measure of how much
energy a charge carries with it. (A 12 V battery gives electrons
12 J of energy for every Coulomb of charge. 1 Coulomb is 6.25
x 1018 electrons). 1 V = 1 J/C
(A scalar)
I
V is the electric potential at a point in an electric field if 1 J of
work is required to move 1 C of charge from infinity to that
point.
V = Ee
q2
V = kq1
r
Electric Potential
 Note:
Any formulae with k in it deals with point charges
(individual charges of small radii) so these can NOT be
used for problems with parallel plates as there are
millions of charges on the plates.
 Potential
difference is the change in electric potential
energy of charge q moving from point A to B (like across
the terminals of a battery might yield a PD = +12 V but
across a resistor the PD might be = - 12 V).
V = E/q
Electric Potential and Parallel Plates
 For
parallel plates, an applied voltage means an
electric field exists between the plates. The plates have
the ability to do work on charges.
 The
parallel plates have an electric field based on the
accumulated effect of millions of charges on the plates:
|E| = V/d
where d is the plate separation
Energy Conservation Problems
 Energy
conservation can be applied such that electric
charges accelerated by electric fields will undergo a
change in kinetic energy.
Examples:
An electron is accelerated by 500. V. a) What is the kinetic energy
gained by the electron in Joules and electron-Volts?
Ee = Ek = qV
= 1.60 x 10-19 C (500 V)
= 8.00 x 10-17 J
8.00 x 10-17 J/(1.60 x 10-19 J/eV)
= 500. eV
Energy Conservation Problems
b) Find the speed of the electron at the end of the acceleration
period.
Ee = Ek
8.00 x 10-17 J = (0.5)mv2
= (0.5)(9.11 x 10-31 kg)v2
v = 1.33 x 107 m/s
Example 2)
An electron is between two parallel plates with the upper plate
positive. An applied voltage of 400. V is attached to the plates and
the plates are 3.00 cm apart. Find the acceleration of the electron.
Potential at a point
Example 3)
3.0 cm
4.0 cm
20.0 nC
 Find
-18.0 nC
J
Ee between the two charges and VT at point J.
Ee = kq1q2
r
= 8.99x109Nm2/C2(20.0 x10-9C)(-18.0 x 10-9C)
(0.040 m)
= - 8.1 x 10-5 J
Potential at a point
Example 3)
4.0 cm
20.0 nC
3.0 cm
-18.0 nC
J
V = kq and VT = V1 + V2
r
= 8.99x109Nm2/C2{(20.0x10-9C) +(-18.0x10-9C)}
{(0.04 + 0.03)m
= 2.569 x 103 V + (- 5394) V
= - 2.8 x 103 V
0.03 m}
Example 4
Find the electric potential energy between a magnesium
nucleus and an alpha particle separated by 12.0 nm.
[4.60 x 10-19 J]