Electric Potential – Show Me The Physics Website

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I.)

Electric Potential –

work needed

or energy acquired

when a positive charge is moved from infinity to a certain point in an

electric field.

  

 

 

II) Potential Difference

(Voltage)

potential energy difference between 2 points in an electric field per

unit of charge

 

 

 

 

A) Potential

Difference, Energy & Work

As a charge moves through an

electric field it gains PE or KE

 

 

 

If “Opposites” move

Closer

 

Gain KE or PE?

 

 

 

 

 

They Gain KE

 

 

 

 

 

 

If ‘Opposites’ Separate

Gain KE or PE?

 

 

 

 

They Gain PE

 

 

 

 

 

 

If “Likes” Separate

Gain KE or PE?

 

 

 

They Gain KE

 

 

 

 

 

 

If ‘Likes’ move closer

 

Gain KE or PE?

 

 

 

They Gain PE

 

 

 

1) Equation #1

 

V =
W

(Joules)

q

(Coulombs)

W

= work done

against field or
energy acquired

working with field (Joules or eV)

 

eV = electronVolts

(small energy unit)

 

 q

– amount of charge moving through field (Coulombs)

V

– Potential Difference (volts)

AP Physics

Charge q moving between 2 points in a field

UE = PE (W)

V =

UE

(Joules)

q

(Coulombs)

UE = qV =

UE =
kq1q2

r

V =

Kq

r

V – electric potential

(Scalar)

 

 

Ex)

It takes 6 J of work to move 2 C of charge between 2 points in an

electric field.

 

What is the potential energy difference (V)

between these points?  

 

Ex) It takes

6 Joules of work

to move 2 coulombs

of charge

between 2 points in an electric field. What is the
potential energy difference

(voltage) between these 2 points?

 

 

V =
W

(Joules)

q

(Coulombs)

 

V

=

6

J

oules

2

C

oulombs

 

 

V

= 3 Volts or J/C

 

 

 

2)
Potential difference Eq. #2

 

 

V =

W

(eV)

q

(elementary

charges)

 

W = work done against field

or energy acquired working with field

 

 

Electronvolt
 � small energy unit

= 1.6 x 10-19 J

 

 

 

q – amount of

charge moving through field (# elementary charges)

 

V – Potential Difference

(volts)

 

Ex) It takes 10

eV to move 2 elementary charges from
one point to another in an electric

field.
Find the potential difference between these points.

 

 

Ex) It takes

10 eV to move

2 elementary charges

from one point to another in an electric field. Find the
potential difference

between these points. 

 

 

V =

W

(eV)

q

(elementary

charges)

 

 

V

=

 

10

e

V

2

e

lementary

charges

 

V = 5 Volts

 

 

 

 

Ex) +2

elementary charges acquire

16 joules of energy when they move between 2

points in an electric field.

What is the

potential difference (voltage)

between these two points in the field?  

 

Ex)
+2 elementary charges

acquire 16 joules

of energy when they move between 2 points in an electric field. What is

the potential difference

(voltage) between these two points in the field?

 

elementary charges, eV  

Joules/Coulombs

 

+2 elementary charges =

3.2 x 10-19 C

 

V =
W

(Joules)

q

(Coulombs)

 

 

V =

16 J

(3.2 x 10-19

C)

 

 

 

 

 

V = 5.0 x 1019

Volts

 

or J/C

 

 

Ex) If 4.8 x 10-17

joules of work is required to move an electron between 2 points

in an electric field, what is the electric potential difference

between these points?

 

a) 3.0 x 102 V
b) 4.8 x 10-17 V

c) 1.6 x 10-19 V
d) 4.8 x 102 V

 

 

 

Ex) If

4.8 x 10

-17

joule of work

i

s required to move an

electron

between 2 points in an

electric field, what is the

electric

potential difference

between these points?

a) 3.0 x 102 V
b) 4.8 x 10-17 V

c) 1.6 x 10-19 V
d) 4.8 x 102 V

 

 

 

W = 4.8 x 10-17

Joule

V = ?

q = ?

 

 

 

 q = 1.6 x 10-19

C

 

 

V =
W

(Joules)

q

(Coulombs)

 

 

V = 4.8 x 10-17 J / (1.6 x 10-19C)

 

 

a) 3.0 x 102 V

 

 

 

AP Physics

 

Electric Field – Parallel Plates

E = V/d = F/q

 

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 Conducting plates are 0.04 m apart, as shown

above.  Point P is located 0.01 m above the lower plate.

 

a. The electric potential at point P is

(A) 10 V (B) 8 V

(C) 6 V (D) 4 V (E)  2 V

 

E = V/d = 8 V/.04m

= 200 V/m

200 V/m = V/.01m

V = 2 V

(A) 10 V (B) 8 V

(C) 6 V (D) 4 V
(E)  2 V

b. The magnitude of the electric field at point P

is

(A)  800 V/m
(B)  600 V/m         

(C)  400 V/m
(D) 

200 V/m         

(E)  100 V/m

E =V/d = 2 V

        

Capacitance

 

Capacitor –

a) Oppositely charged plates connected to a voltage

source

b) used to store charge

c) once capacitor is charged, voltage source can be

removed

 

Q = CV

Q – charge

V – Voltage

C – Capacitance – Farads

 

A = Area of each plate

d – distance between plates

E0 = epsilon (ref)

= 8.84 x 10-12 C/Vm

 

 

Millikan

 

 

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