7/25/2019 P102 M1 Summer 15-16 Sol

1/3

PHYS 102 General Physics IISummer 2015 16, Midterm Exam 1 Solutions11 June 2016, 13:30

1. Two point charges are placed on axis as shown inthe figure. The first one has a charge +2 andcoordinate ( ,0). The second one has a charge and coordinate (, 0). Please use vector notation toanswer the following questions, that is, if you have a

vectorit should be written as= + + .

(a) (8 Pts.) What is the force on the first particle due to

the second particle ( ) and vice versa ( )?Solution:

2

12 2 2

0 0

(2 )( ) .4 (2 ) 8

Q Q Qa a

F i i

2

21 12 2

0

.

8Qa

F F i

(b) (9 Pts.) What is the total electric field along the line

connecting the two point charges (as a function of )?Solution:

1 2

0

2

4 ( )

Q

a x

E i , 2 20

4 ( )

Q

a x

E i .

2 2

1 2 22 2

0

3 3 2.

4

Q a x ax

a x

E E E i

(c) (9 Pts.) What is the total electric field along the

axis (as a function of )?Solution:

1 2 2

0

2

cos sin4 ( )

Q

a y E i j ,

2 2 20

cos sin4 ( )

Q

a y

E i j ,2 2 2 2

cos , sina y

a y a y

1 2 2 2 3/ 20

34 ( )

Qa y

a y

E E E i j .

(d) (8 Pts.) Draw the electric field lines. Comment on the meaning of your drawing and its relationship to

your answers in part (b) and (c).

-Q2Qx

y

.x

+

-Q2Qx

y

y

+ +

a a

7/25/2019 P102 M1 Summer 15-16 Sol

2/3

(d) (8 Pts.) Solution:

The electric field lines along the axis are closer together in the left hand side compared to the right handside. This is in accordance with the answer in part (b). The electric field lines along the axis are

pointing towards right with a little contribution in direction. This is in accordance with the answer inpart (c).

2.(a) (10 Pts.) A uniform field E is parallel to the axis of a hollow hemisphere ofradius r. What is the electric flux through the hemispherical surface?

Solution:

From the figure, we see that the flux outward through the hemispherical

surface is the same as the flux inward through the circular surface base of the

hemisphere. On that surface all of the flux is perpendicular to the surface. So, we say that on the circular base,.

E A

ThusE

2.r E E A

(b) (23 Pts.) Two thin concentric spherical shells of radii1r and

2r 1 2r < r contain uniform surface charge densities

1 and

2, respectively. Determine the electric field for

10 ,< r < r

1 2,r < r < r and

2.r > r Under what conditions will 0E for

2?r > r Neglect the thickness of the

shells.

Solution:

In the region1

0 ,r r a gaussian surface would enclose no charge. Thus, due to the spherical symmetry, we have

7/25/2019 P102 M1 Summer 15-16 Sol

3/3

2 encl0

4 0 0.Q

d E r E

E A

In the region21

,r r r only the charge on the inner shell will be enclosed. Therefore

2

2 encl 1 1

0 0

2

1 1

2

0

44 .

Q rd E r E

r

r

E A

In the region2

,r r the charge on both shells will be enclosed. So

2 2

2 encl 1 1 2 2

0 0

2 2

1 1 2 2

2

0

4 44 .

Q r rd E r E

r r

r

E A

To make 0E for2

,r r we must have2 2

1 1 2 2 0.r r This implies that the shells are of

opposite charge.

3.The electric potential in a region of space is given by (,,)= (2 3+ ),where is in volts, the coordinates are in meters, andis a constant.

a) (11 Pts.) Find the electric field in this region (express the result in terms of unit vectors).Solution:

x y zE E E E i j k , where 4 , 6 , 2x y z

V V VE Ax E Ay E Az

x y z

.

Hence

2 2 3A x y z E i j k .

b) (11 Pts.) If the work done by the field on a 4-Ctest charge when it moves from point ,0,1mto the origin(0,0,0)is 3.105J, what is? (: 1 = 106)

Solution:

0 0 ( )E f iW U q V q V V .21

2

3( , 0,1) , (0, 0, 0) 0

2i f

AV V m V V . Since 6

0 4 10q C ,

5

0 0

33 10

2EW q V q A J

and we have

25 /A V m .

c) (11 Pts.) Find the magnitude of the electrostatic force acting on the test charge at the point ,0,1m.Solution:

12

( , 0,1) 10( ) /V m E i k 10 2 /E V m .5

0 4 2 10F q E N

.