Exam 2 Practice Problems Part 1 Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

Exam 2 Practice Problems Part 1 Solutions

Problem 1 Electric Field and Charge Distributions from Electric Potential An electric potential V (z) is described by the function

)(!2V " m-1)z + 4V ; z > 2.0 m

++0 ; 1.0 m < z < 2.0 m

V

(

z)

=

*+++++++02233;VV+!!#$%2#$%.23203

V

"

m

-3

& '(

z3;

0

m

<

z

<

1.0

m

V

"

m-3

& '(

z3;

!

1.0

m

<

z

<

0

m

m < z < ! 1.0 m

,+(2V " m-1)z + 4V ; z < ! 2.0 m

The graph below shows the variation of an electric potential V (z) as a function of z .

! a) Give the electric field vector E for each of the six regions in (i) to (vi) below?

Solution:

We

shall

the

fact

that

! E

=

!"! V

.

Since

the

electric

potential

only

depends

on

the variable z , we have that the z -component of the electric field is given by

Ez

=

!

dV dz

.

The electric field vector is then given by

(i) z > 2.0 m :

! E

=

Ez

^i

=

!

dV dz

k^

( ) !

E

=!

dV

k^

=

!

d

(!2V " m-1)z + 4V k^ = 2V " m-1 k^

dz

dz

(ii) 1.0 m < z < 2.0 m : (iii) 0 m < z < 1.0 m:

! E

=

!

dV

k^

=

! 0

dz

( ) !

E

=

#

dV dz

k^

=

#

d dz

! %'

2 3

V

# !%'

2 3

V $ m-3

"&(

z3

" &(

k^

=

2 V $ m-3

z 2k^

Note that the z2 has units of [m2 ], so the value of the electric field at a point just inside z- = 1.0 m - m where > 0 is a very small number is given by

E! ! = (2 V " m-3)(1 m)2 k^ = 2 V k^ m

Note that the z-component of the electric field, 2 V ! m"1 , has the correct units. (iv) -1.0 m < z < 0 m :

( ) !

E

=

!

dV dz

k^

=

!

d dz

# $%

2 3

V+

# $%

2 3

V

"

m-3

& '(

z

3

& '(

k^

=

!

2 V " m-3

z2 k^

The value of the electric field at a point just inside z+ = !1.0 m + " m is given by

E! !

=

!(2

V

" m-3)(1 m)2

k^

=

!2

V m

k^

.

(v) -2.0 m < z < -1.0 m:

! E

=

!

dV

k^

=

! 0

dz

(vi) z < ! 2.0 m :

( ) !

E

=!

dV

k^

=

!

d

(2V " m-1)z + 4V k^ = !2V " m-1 k^

dz

dz

b) Make a plot of the z-component of the electric field, Ez , as a function of z . Make sure you label the axes to indicate the numeric magnitude of the field.

c) Qualitatively describe the distribution of charges that gives rise to this potential landscape and hence the electric fields you calculated. That is, where are the charges, what sign are they, what shape are they (plane, slab...)?

In the region !1.0 m < z < 1.0 m there is a non-uniform (in the z-direction) slab of positive charge. Note that the z-component of the electric field is zero at z = 0 m , negative for the region !1.0 m < z < 0 m , and positive for 0 m < z < 1.0 m as it should for a positive slab that has zero field at the center. In the region 1.0 m < z < 2.0 m there is a conductor where the field is zero. On the plane z = 2.0 m , there is a positive uniform charge density ! that produces a constant field pointing to the right in the region z > 2.0 m (hence the positive component of the electric field). On the plane z = 1.0 m , there is a negative uniform charge density "! . In the region !2.0 m < z < !1.0 m there is a conductor where the field is zero. On the plane z = ! 2.0 m , there is a positive uniform charge density ! that produces a constant field pointing to the left in the region z < ! 2.0 m (hence the positive component of the electric field). On the plane z = !1.0 m , there is a negative uniform charge density "! .

Problem 2: Electric Field from Electric Potential The electric potential V (x) for a planar charge distribution is given by:

) +

0

+ V (x) = *++

+ + +

!V0

"#$ 1 +

x d

% &'

2

!V 0

" #$

1

+

2

x d

% &'

,+

!3V 0

for x < !d for ! d ( x < 0

for 0 ( x < d for x > d

where !V0 is the potential at the origin and d is a distance.

This function is plotted to the right, with d = 2 cm and V0 = 2 V , the x-axis with units in cm, the y-axis in units of Volts.

! (a) What is the electric field E(x) for this problem?

Region I: x < -d

Region II: !d " x < 0

! E

=

!"V

=

!

#V #x

^i

=

0

! E

=

!

" "x

# $%%

!V0

# $%

1

+

x d

& '(

2& '((

^i

=

2 V0 d

#$%1 +

x d

& '(

^i

Region III: 0 ! x < d Region IV: x > d

! E

=

!

" "x

# $%

!V0

#$%1

+

2

x d

& '(

& '(

^i

=

2 V0 d

^i

! E

=

!"V

=

!

#V #x

^i

=

0

(b) Plot the electric field that you just calculated on the graph below. Be sure to properly label the y-axis (top and bottom) to indicate the limits of the magnitude of the E field!

This makes sense. Where the potential is flat, the E field is 0.

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