Electric Force, Field & Potential

A comparison

Example Three charges, q1 = +8 C, q2 = +6 C and q3 = -4 C are arranged as shown below. Find the

resultant force on the –4 C charge due to the others.

Draw free-body diagram.Draw free-body diagram.

-53.1o

-4 Cq3

F1

F2

Note the directions of forces FNote the directions of forces F11and Fand F22 on on qq33

based on attraction/repulsion from based on attraction/repulsion from qq1 1 and and qq22. .

+ -

4 cm

3 cm

5 cm

53.1o

+6 C

-4 C

+8 Cq1

q2

q3

+

Example (Cont.) Next we find the forces F1 and F2 from Coulomb’s law. Take data from the

figure and use SI units.

9 -6 -6

1 2

(9 x 10 )(8 x 10 )(4 x 10 )

(0.05 m)F

F1 = 115 N, 53.1o S of WF1 = 115 N, 53.1o S of W

9 -6 -6

2 2

(9 x 10 )(6 x 10 )(4 x 10 )

(0.03 m)F

F2 = 240 N, WestF2 = 240 N, West

1 3 2 31 22 2

1 2

; kq q kq q

F Fr r

Thus, we need to find resultant of two forces:Thus, we need to find resultant of two forces:

+ -

4 cm

3 cm

5 cm

53.1o

+6 C

-4 C

+8 Cq1

q2

q3

F2

F1

+

Example 4 (Cont.) We find components of each force F1 and F2 (review vectors).

53.1o- -4 C

q3

F1= 115 N

F1y

F1x

FF1x1x = - = -(115 N) Cos 53.1(115 N) Cos 53.1oo

= - 69.2 N= - 69.2 N

FF1y1y = - = -(115 N) Sin 53.1(115 N) Sin 53.1oo = =

- 92.1 N- 92.1 NNow look at force FNow look at force F22::

FF2x2x = = -240 N; F-240 N; F2y2y = 0 = 0 RRxx = =FFxx ; R ; Ry y = = FFyy

RRxx = – 69.2 N – 240 N = -309 N = – 69.2 N – 240 N = -309 N

RRyy = -69.2 N – 0 = -69.2 N = -69.2 N – 0 = -69.2 N

F2

240 N

Rx= -92.1 NRx= -92.1 N

Ry= -240 NRy= -240 N

Example 4 (Cont.) Next find resultant R from components Fx and Fy. (review vectors).

Rx= -309 NRx= -309 N Ry= -69.2 NRy= -69.2 N

- -4 Cq3

Ry = -69.2 N

Rx = -309 N

R

We now find resultant R,We now find resultant R,::

y2 2

x

R; tan =

Rx yR R R y2 2

x

R; tan =

Rx yR R R

2 2(309 N) (69.2 N) 317 NR

R = 317 NR = 317 NThus, the magnitude of Thus, the magnitude of the electric force is:the electric force is:

Example (Cont.) The resultant force is 317 N. We now need to determine the angle or

direction of this force.

2 2 317 Nx yR R R 2 2 317 Nx yR R R

y

x

R 309 N tan

R -69.2 N

-69.2 N

--309 N

R

-62.9 N

Or, the polar angleOr, the polar angle is: is: = 1800 + 77.40 = 257.40

The reference angleThe reference angle is: is: = 77.40S of W

Resultant Force: R = 317 N, = 257.40Resultant Force: R = 317 N, = 257.40

Fields are vectors, too1. Now, consider point P a 1. Now, consider point P a

distance distance rr from +Q. from +Q.

2. An electric field E exists 2. An electric field E exists at P if a test charge +q at P if a test charge +q has a force F at that point.has a force F at that point.

3. The direction of the E is 3. The direction of the E is the same as the direction of the same as the direction of a force on + (pos) charge.a force on + (pos) charge.

E

4. The magnitude of E is 4. The magnitude of E is given by the formula:given by the formula:

N; Units

C

FE

q

N; Units

C

FE

q

Electric Field

++++ +

+++Q

.P

r

+qF

+

The E-Field at a distance r from a single charge Q

++++ +

+++Q

.r

P

Consider a test charge +q placed Consider a test charge +q placed at P a distance r from Q.at P a distance r from Q.

The outward force on +q is:The outward force on +q is:

The electric field E is therefore:The electric field E is therefore:

2F kQq rE

q q 2

kQE

r 2

kQE

r

++qF

2

kQqF

r

++++

+

+++Q

.r

P

E

2

kQE

r

The Resultant Electric Field.The resultant field E in the vicinity of a number The resultant field E in the vicinity of a number of point charges is equal to the vector sum of the of point charges is equal to the vector sum of the fields due to each charge taken individually.fields due to each charge taken individually.

Consider E for each charge.Consider E for each charge.

+

- q1

q2q3

-

AE1

E3

E2

ER Vector Sum:

E = E1 + E2 + E3

Vector Sum:

E = E1 + E2 + E3

Directions are based on positive test charge.

Magnitudes are from:

2

kQE

r

Electrical Potential, Work and Energy

An external force F moves +q from A to B against the field force qE.

Work = Fd = (qE)dAt level B, the potential energy U is:

U = qEd (Electrical)

The E-field does negative work; External force does positive work.

The external force F against the E-field increases the potential energy. If released the field gives work back.

The external force F against the E-field increases the potential energy. If released the field gives work back.

B + + + +

- - - -A

++q d

qE E

Fe

Work to Move a Charge

++++ +

+++Q

qE

F

Work to move +q from Ato

B.

ra

rb

2aa

kqQF

r

avga b

kqQF

r r

At A:

At B:

Avg. Force:

2bb

kqQF

r

Distance: ra - rb

( )a ba b

kQqWork Fd r r

r r 1 1

b a

Work kQqr r

1 1

b a

Work kQqr r

Electric Potential

Potential

++++ +

+++Q

.r

Electric potential is another Electric potential is another property of space allowing us to property of space allowing us to predict the P.E. of any charge q predict the P.E. of any charge q at a point.at a point.

UV

q

; U

V U qVq

; U

V U qVq

Electric Electric PotentialPotential::

The units are: joules per coulombjoules per coulomb (J/C)(J/C)

For example, if the potential is 400 J/C at point P, For example, if the potential is 400 J/C at point P, a –2 nC charge at that point would have P.E. :a –2 nC charge at that point would have P.E. :

U = qV = (-2 x 10-9C)(400 J/C); U = -800 nJU = -800 nJ

P

Calculating Electric Potential

Potential

++++ +

+++Q

.r

kQV

rP

Electric Potential Energy and Electric Potential Energy and Potential:Potential:

; kQq U

U Vr q

; kQq U

U Vr q

kQqr kQ

Vq r

Substituting for Substituting for U, we find V:U, we find V:

kQV

r

kQV

r

The potential due to a positive charge is positive; The potential due to a negative charge is positive. (Use sign of charge.)

The potential due to a positive charge is positive; The potential due to a negative charge is positive. (Use sign of charge.)

Potential For Multiple Charges

The Electric Potential V in the vicinity of a number of charges is equal to the algebraic sum of the potentials due to each charge.

The Electric Potential V in the vicinity of a number of charges is equal to the algebraic sum of the potentials due to each charge.

+

- Q1

Q2Q3

-

Ar1

r3

r2

31 2

1 2 3A

kQkQ kQV

r r r

kQV

r

kQV

r

Potential is + or – based on sign of the charges Q.Potential is + or – based on sign of the charges Q.