1516 Level NS Core Physics BGT Questions Solutions Electricity

Physics Core Course Basic Questions Table of Contents

I.

ELECTRICITY: 1. 2. 3. 4. 5. 6. 7. 8.

Electric Charge And Electric Field Gauss’s Law

Electric Potential Capacitance and Dielectrics Current, Resistance and E.M.F. Direct-Current Direct-Current Circuits Magnetic Fields and Magnetic Forces Sources of Magnetic Fields

1 5 10 13 17 20 24 27

II.

RELATIVITY:

31

III.

Mechanics Part 1 Mechanics Part 2 Mechanics Part 3 Mechanics Part 4 Mechanics Part 5 Mechanics Part 6 Mechanics Part 7 Mechanics Part 8

35 38 41 50 53 57 60 62

IV.

Chapter 21 (Student Kinematics) Kinematics )

65

V.

WAVES: Chapter 15 (Mechanical Waves) Chapter 16 (Sound and Hearing)

71 76

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Chapter 21: Electric Charge and Electric Field

B 1. (a)

State Coulomb’s law and give its mathematical expression. [chap 21-page 717]

Solution:

The magnitude of the electric force between two point charges is directly proportional to the product of the charges charges and is inversely proportional to the square of the distance between them.

F (b)



k q1q 2 r 2

Two point charges, q1 = -3.0 nC and q 2 = +5.0 nC are separated by a distance of 2.0 cm. Find the magnitude of the electric force that q1 exerts on q 2. [chap 21-page 720]

Solution:

(c)

F 

k q1q 2 2

r



9 9 k  3.0  10 10   5.0  10 10 

 2 m  2 . 0cm 1 0    cm  

2

 3.37  104 N

Define electric field. [chap 21-page 722]

Solution:

B 2. (a)

The electric field is defined as the electric force F experienced by a test charge q0 placed in the field divided by the charge q0 (providing q0 is small enough so as not to disturb the charges generating the field).

Find the magnitude of the electric field at a field point 4.0 m from a point charge q = 2.0 nC. [chap 21-page 724]

Solution:

E  E 

kq r 2

k 2. 2.0 0  10 9 



(4.0)2



1.12 1.1 2 N/ N/C C















Solution:

E  E



kq 2

q



E r 2

(2..0cm 10 2 m/c (2 /cm m) 2  4.5 10 5  







2.0 10 8 C 

r k k Since the field is away from the charge, the charge is positive 8

q  2.0 10

B 3.

(a)

C

Find the distance from a charge of 16 16 nC where the electric field due to the charge has 4 -1 a magnitude of 4.0 x 10 NC .

Solution:

E 

kq r

2

k 16 10 

9





2

r

k .(16 10 9 )

4.0 109





(b)

r 

4.0 10

4



6.00 10 2 m 

A point charge q = 6.00 nC is located at the origin. origin. What is the magnitude of the electric field vector at the point of coordinates (- 15.0 cm, - 20.0 cm)? [chap21-page 725]

Solution:

Since q is positive, the electric field is away from it. + y Q = 6.00 nC – 15 15 P

E 



B 4.

kq r 2

+ x – 20 20

9



k 6. 6.00 00 10 



(0.1 (0 .15 5)

2



(0.2 (0 .20) 0)

2



2

E  8.63 10 2 N/C 

Three charges lie along the x-axis x-axis as shown in the diagram. The positive charge q1 = 4.00 C is at x at x = 4.00 m, and the positive charge q2 = 8.00 C is at the origin. A negative charge q3 is placed on the x-axis x-axis such that the resultant force on it is zero. What is the x the x-coordinate -coordinate of q3? O q3 q2

q1 [chap 21-page 720]















 F x  

0  F13

k . q1 q3 (r13 )





q1

(r13 )2 2



k . q2 q3



2

F 23



( r 23 ) 2

q2

(r23 )2 x



8 106 x

2



4 10 6 (4  x) 2

2 2

16  8 x  x

2

 32  16 x  2 x  x

2

2

 x  16 x  32  0 2

2

  b  4ac  ( 16)  4(1)(32)  128  x 

b 



16  128

 x  13.66 or x  2.34 2a 2 x is x is a distance so it must be positive, x is also between 0 and 4 as given in the figure:  x 

16 

x

128 

2

2.34m 2. 34m

BG 5. y q2 -

a

+ q3

a q1 +

x

Consider three point charges located at the corners of a triangle, as shown in the diagram, where q1 = q3 = 3.00 C, q2 = - 8.00 C, and a = 0.120 m.

G

(a)

Find the magnitude of the force F1 of q1 on q3.

(b)

Find the magnitude of the force F2 of q2 on q3.

(c)

What is the x-component of the resultant force on q3?

(d)

What is the y-component of the resultant force on q3?

(e)

Find the magnitude of the resultant force on q3.

(f)

Calculate the angle the resultant force makes with the positive x-axis. [chap 21-page 721]



















tan

1 a

tan





1



1



45

0

a

(a)

F 1

k 3.00  10

k q1 q3





(r 13 ) 2



a

6



2

3.00 10

k 8.00 10

k q2 q3

a



6



2



6





2

3.00 10

2.81N

6



(b)

F 2

(c)

R x  F1x  F2 x  F1 cos 45 i  F2 x i  2.81cos 45i   15i   13.01i N

(d)

R y



(e)

R



(f)

  tan 1





(r23 )2

a2



F1 y



2

R x

F2 y



2

Ry







 

15 N





 F1 sin 45 i  0  1.99i N

13.16 .16 N  13

 1.99  -axis  tan 1   8.70 with the positive x-a  R x 13.01  

R y

B 6.

P

+

x















(a)

Find the magnitude of the electric field at P due to q1.

(b)

Find the magnitude magnitude of the electric field at P due to q2.

(c)

What is the x-component of the resultant electric field at P?

(d)

What is the y-component y-component of the resultant electric field at P?

(e)

Find the magnitude of the resultant electric field at point P.

(f)

Calculate the angle the resultant electric field makes with the positive x-axis. [chap 21-page 728]

Solution:

Since q1 is positive, the electric field at P due to q 1 is away from q 1. Since q2 is negative, the electric field at P due to q2 is towards q 2.

  tan

(a)

E 1

(b)

E 2

(c)

E x

0.6 

(r 1 )



2

k q2 (r 2 )

E1x

tan

0.8

k q1





1

2



1

3 

36.87

0

4

k 8. 8.00 0010 6 



 0.6

2



1.99 1. 99 105 N/C N/C

k 3. 3.00 00106 



E2 x

0.62  0.82







2.70 2. 70 104 N/C

0  E2 cos  i  2.16 104 N/C i

















(f)

E  y  1 y   83.30 above the +ve x-a t a n -axis  E x  x 

  tan 1 

BG 7.

y

l

x

--------------------

v0 E

+ +| + + + + + + + + + + + An electron of mass 9.11 x 10 -31 kg enters the region of a uniform electric field as shown in the diagram, with v 0 = 2.00 x 10 6 m/s and E = 300 N/C. The width of the plates is l = l = 0.250 m. G

(a)

Find the acceleration of the electron while in the electric field.

G

(b)

Find the time it takes the electron to travel through the region of the electric field.

(c)

What is the vertical displacement y of the electron while it is in the electric field?

(d)

Calculate the speed of the electron as it emerges from the electric field.

G

[chap 21-page 726] Solution:

Since the electron e – is negatively charged, the electric force on it is in a direction opposite to the direction of the electric field.

















t 

(c)

X

2

at

  y 

(d)

6

2.00 2. 00 10

V 0

1

 y 

0.250



2

1 2

V y



V x

 Vox 

 Voyt 

1 2

 t  1.25  10

2

at

7

0

(5.27 10 1013 j )(1.25 10 107 ) 2

at  Vo



s

 0.412

mj

(5.27  1013 )(1.25 10 7 )  0   6.59 106 m/s j 

6

2.00 10 m/si

 V   V  V x2  V y2  6.88 106 m/s;  tan 1  y   73.10  V x  6.88 88 106 m/s at an an angl glee of of 73. 73.1 10 be belo low w th thee +ve +ve xx-ax axis is  V  6.

B 8.

(a)

In a thundercloud there may be an electric charge of + 30.0 C near the top and – and – 30.0 30.0 C near the bottom. These charges are separated by approximately 3.00 km. Find the magnitude of the electric force between them. [chap 21-page 720]

Solution:

(b)

F



k q1 q2 2



k3 0

r

30 3 2

(3 10 )

5

.99 9 10  8.9

N

What are the magnitude magnitude and direction of the electric field that will balance the weight weight -27 of three protons, of mass 1.67 x 10 kg each? [chap 21-page 725]

Solution:

+y

 F y  0 

Fe



E q0



E

mg





Fe



m.g

3.m proton .g 3

q proton



3 1.67  10 10 



  9.8 10 

27

3 1.60 

19

 1.02  10

7

N/C

mg

















+y  F y  0  T  Fe   

(b)

T

mg

Fe

T 

mg  Fe

T 

(0.045)(9.8)  E

 T 

(0.0 (0 .04 45)(9. 9.8 8)  ( 2. 2.0 00 10 3 )( 40. 0.0 0 10 6)

 T 

0.361N



  q0 

E

Q = – = – 40.0 40.0 C 

The applied applied electric field holds the sphere in place above the fixed point of suspension, and the tension in the thread is 0.450 N. Find the magnitude of the electric field. [chap 21-page 722]

Solution:

Since Q is negative and F e is vertically upwards, E must be vertically downwards: +y  F y  0  Fe    T  0  E.q 0 j  ( mg ) j  (0.450) j Fe (9.8 (9 .8)(0 )(0.04 .045) 5) j  (0. (0.450 450)) j  E  40.0 .0 106 40 Q = – = – 40.0 40.0 C 4

.23 3 10  E  2.2 

E



N/C j

2.2 2. 23 10 4 N/C j

 T

















Chapter 22: Gauss’s Law

BG 1. G

(a)

State Gauss’s law in words and give its mathematical expression. [chap 22-page 759]

Solution:

The total electric flux through t hrough any closed surface is directly proportional to the net electric charge enclosed within the surface.  E 

(b)

 E.dA

When excess charge is placed on a solid conductor, where does it reside? [chap 22-page 761]

Solution:

G

(c)

Entirely on the surface.

Write the three forms of the general definition of electric flux. [chap 22-page755]

Solution:

G

(d)

 E 

 E dA   E .

 .dA

  E cos dA

Give the expression of the magnitude of the electric field outside a charged conducting sphere. [chap 22- page 762]

Solution:

















BG 2. G

(a)

What Gaussian surface is used to find the the electric field outside an infinitely long long charged wire? [chap 22-page 763]

Solution:

G

(b)

A cylinder with the wire as its it s axis and with ends perpendicular to the wire. The radius of the cylinder is R and the length is L.

What is the magnitude of the field of an infinite sheet of charge? [chap 22-page 764]

Solution:

Take as Gaussian surface a cylinder bisected by the sheet of charge. E=

E  since E points

perpendicularly away from the sheet at ever y point.



 E  E.dA but  E  1  2  3 ; 3  0 since E   0

  1   E.dA   E .dA  E  dA  EA







&  2  E.dA  E .dA  E dA  EA

  E  1   2  2 EA   2 EA 

 . A 0

 E 

E

1 E

2 0

 (c)

 0





G

3

qenc

+ + + + + + + + + + +

E

2 E

Surfaces 1 & 2 are the ends of the cylinder; surface 3 represents the lateral area of the cylinder. q Surface charge density =  A 

What is the magnitude of the field between oppositely charged parallel conducting plates? [chap 22-page 765]




















Induced charge on inner surface = +2.0 nC

B 3. S!

S3

S2 q2 q4 q1

q3 q5

S4 Four closed surfaces S 1 through S 4, together with the charges q 1, q2, q3, q4 and q5 are sketched in the figure. figure. q1 = -4Q, q2 = +Q, q3 = -2Q, q4 = -3Q, q5 = -Q (a)

Find the electric flux through surface S1.

(b)


(c)


(d)

Find the electric flux through surface S4. [chap 22-page 759]



















(b)

The total electric flux through through a closed surface in the shape of a cylinder is 4 4.80 x 10 Wb. What is the net charge within the cylinder? (permittivity of free space = 8.85 x 10 -12) [chap 22-page 759]

Solution:

 E



qenc



 0

qenc

 E . 0 

 4.8 10     4 .25 10 4

0

7

C

conducting plate 35.0 cm on a side lies in the xy plane. A total charge of B 5. A thin square conducting 1.96 x 10 -8 C is placed on the plate. (permittivit y of free space = 8.85 x 10 -12). (a)

Find the charge density on the plate.

(b)

What is the magnitude of the electric field just above the plate? [chap 22-page 764]

Solution:

Since the plate is thin, we may model it as a sheet. (a)

 

q A

8



1.96 1. 96 10

(0.35)(0.35)

 1.6 10

7

C/m C/ m2

1.6 10  7

(b)

for a thin plate (sheet):

E 



2 0



2 0

3

 9.04 10

N/C

conducting spherical shell of radius 13.0 cm carries a net charge charge of 5.07 5.07 C uniformly BG 6. A conducting



















(a)

the inner surface.

(b)

the outer surface. [chap 22-page 767]

Solution:

Since Q = +4.00 C in the cavity, a charge of – of – 4.00 4.00 C is induced on the inner surface of the shell. Consequently the charge on the outer surface of the shell is +13.00 mC (since 13 + ( – 4) 4) = 9)

(a)

(b)

B 8.

 inner 

 outer 

qinner

4.00 10

 

6



2

4.00 10

 

6



2

Ainner

4 r

4 (0.05)

qouter

13.00 106

13.00 10 6

Aouter



2

4 r



4 (0.06)

2



1.2 .27 7 10 4 C/m2

.87 7 10  2.8



4

C/m2



















* Note:

(a)

1) the electric field caused by a positively charged plate always points away from it. 2) the electric field caused by a negatively charged plate always points towards it.

At A we must consider the two electric fields

E1 and E 2 caused

by

plates 1 and 2 respectively: Etotal



E1  E 2





 2 0 7

 E A  6.0  10

(b)

2 0



 0

(5.3 (5 .31 110 4 ) 



0



6.0 107 N/C

 N/C i

Both sheets have positive uniform charge densities 1 = 2 = 5.31 x 10 -4 C/m2. What is the electric field at point B?

Solution:

(c)





At B, the two electric fields

E1 and E 2 are

in direction; they cancel so:

E B



equal in magnitude and opposite

0 N/ N/C C.

Both sheets have positive uniform charge densities  1 =  2 = 5.31 x 10 -4 C/m2. What is the electric field at point C?

Solution:

At C, we must consider the two electric fields E1

and

E 2 caused by plates 1

and 2 respectively: Etotal  E1  E 2 



2 0





2 0



 0



5.31 5. 31104 0

6.0 0 107 N/C N/C  6.



















(d)

At A, the two electric fields

E1 and E 2

(caused by plates 1 & 2







2 0

respectively) are equal in magnitude    direction; they cancel so: (e)

E A



  but opposite in 

0 N/ N/C C

The sheet on the left has a uniform surface density 1 = 5.31 x 10 -4 C/m2, and the one on the right has a uniform charge density 2 = -5.31 x 10 -4 C/m2. What is the electric field at point B?

Solution:

E1 and E 2 caused

At B, we must consider the two electric fields

by plates 1

and 2 respectively: Etotal



E1  E 2

 E B 

(f)

1.2





2 0 8

10



 2 0



 0

4





5.31 5. 3110 0



1.2 1. 2 108 N/C N/C

 N/C i

The sheet on the left has a uniform surface density 1 = 5.31 x 10 -4 C/m2, and the one on the right has a uniform charge density  2 = -5.31 x 10 -4 C/m2. What is the electric field at point C? [chap 22-page 765]

Solution:

By a similar argument to that in (d), we can show that:

E C  0 N/C

B 9.

y























5

3

10

    5 2

r



3

10

    0.25

2

  E (left )  312.5 Nm2/C  312.5 b This flux is into the surface, therefore: (b)







 E left

  b

 312.5



 E right  E2 .dA  E  .dA dA E 2 . dA  E 2 A 

1.5





2

103   r 2   1.5 103     0.25

 E (right )  93.75 Nm2 /C  93.75 b

This flux is out of the surface, therefore: (c)

 Etotal



(d)

 Etotal





2

qenc

Eleft qenc  0 



E right  312.5

 qenc  E

6.08 6.0 8 nC

total

. 0



 93.75



 E right



93.75

 b

 b

 218.75

 ( 218.75 )( 0 )  6.08 10

9

C





















the middle sphere is +Q. This induces a charge of – of – Q on the inner surface of the outermost sphere. The net charge on the outermost sphere sphere is – Q, Q, so the charge on the outer surface of the outermost sphere is 0. (a)

+Q

(b)

0





















Chapter 23: Electric Potential

BG 1. G

(a)

What is the electric potential energy of two point charges q and q0 a distance r apart? [chap 23-page 784]

Solution:

(b)

U

1 

4 0

.

qq0 r

A pair of charged metal plates (the top is negative and the bottom is positive) sets up a uniform electric field with magnitude E. What are the direction and magnitude of





















(b)

The potential difference difference between two oppositely charged parallel metal plates is 450 V. Find the work done on a charge of 40 mC as it moves from the higher potential plate to the lower. [chap 23-page 788]

Solution:

(c)

Wab

 U a  Ub  q0 (Va  Vb )  (40 10

3

)(450  0) 0)  18 18J

When a charge of 20 C is moved between two points M and N, in a uniform electric field, 240 J of work is done. What is the potential difference between M and N? [chap 23-page 788]

Solution:

V MN 

W MN q0

6





240 10 20  10

6



 12.0V

BG 3. G

(a)

A positive charge of 4.0 C is moved through an electric field to a point where the potential is +250 V higher than before. Find the increase in the potential energy of the system.





















 V 

BG 4. G

6

2.94 2. 9410 m/s

A 3.00- 3.00-C point charge is located at the origin, and a second point charge of – 6.00 6.00 C is located on the y-axis at the position (0,2.00) m. (a) Find the total electric potential due to these charges at the point P, whose coordinates are (1.50,0) m. (b) What is the work required to bring a -3.00- -3.00-C point charge from infinity to the point P? [chap 23-page 792]

Solution:

q2 =-6.00 C

2.00 m

2.0

2

 1.5

2























Solution:

k.e  q.V

1 

mV

2  V 

2

1 

2

2

mV0

 

1.602

 10

19



105



(1. 1.42 42 105 )m ) m/s

*Note: the proton will accelerate from +ve to – ve: ve:

V = Vfi na  V in ititi al nal al

since V

decreases as the proton moves from +ve to t o – ve: ve: V = = 0 – 0 – 105 105 = – = – 105 105

B 7. An electron, of mass 9.11 x 10 -31 kg, moving parallel to the x-axis has an initial speed of 6

5























Solution:























Chapter 24: Capacitance and Dielectrics

BG 1.

(a)

What is the capacitance of a parallel-plate capacitor? Define all its terms.
























1 Ceq

BG 3.

(a)



1 C1



1 C2



1



1

C3 2.0 10

6



1 4.0 10

6



1 9.0 10

6

Find the equivalent capacitance of the following combination. 2.0 F

 C eq  1.16  10

6

F























(a)

Find the capacitance of the device. [chap 24-page 830]

(b)

What is the maximum energy that can be stored in the capacitor? [chap 24-page 833]

Solution:

(a)

C 

 A d



k 0 A d

0.05   4.90 8.85 10 12   0.015  0. 



3



1.40 1. 40 10



2.32 10 10

11



F

























 F y  0  Ty 

mg  T cos18  (0.3)(9.8)  T  3.09 N























  4.00 106 V1   6.00 10 6  15  V 1 

























Ceq .V0



C1V  C2V



3.2



4



10





V  4  10

6





16.00 10 6  













































G

(b)

A cell of e.m.f. 12.0 V and internal resistance 1.50  is connected across a 2.50- 2.50- resistor. Find the charge which passes any point in the cir cuit in three minutes.







































































































































































































































































1516 Level NS Core Physics BGT Questions Solutions Electricity

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