Faculty of dentistry
Department of Basic Sciences
Date: 9-4-2011
Quiz 1
Time: 1/2h
Student name …………………………
ID...........................................
Circle the correct answer:
1- A conducting sphere has charge Q and its electric potential is V , relative to the potential far away. If the charge is
doubled to 2Q, the potential is:
A. V
B. 2V
C. 4V
D. V/2
E. V/4
2 The units of 1/4πε0 are:
A. N2C2
B. N · m/C
C. N2 · m2/C2
D. N · m2/C2
E. m2/C2
3- The diagram shows the electric field lines in a region of space containing two small charged spheres (Y and Z). Then:
A. Y is positive and Z is negative
B. the magnitude of the electric field is the same everywhere
C. the electric field is strongest midway between Y and Z
D. Y is negative and Z is positive
E. Y and Z must have the same sign
4- The units of resistivity are:
A. ohm
B. ohm·meter
C. ohm/meter
5- Energy may be measured in:
A. kilowatt
B. joule·second
D. ohm/meter2
C. watt
E. none of these
D. watt·second
E. volt/ohm
Answer only one problem:
1 Calculate the speed of an electron that is accelerated from rest through a potential difference of 90 V.
V=90 V
,me=9.1x10-27 Kg
qV=1/2mv2
,q=e=1.6x10-19C
1.6x10-19 x 90 =1/2 9.1 x 10-27 v2
v = 5.6 x 106 m/s
2- Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 10.0 g. Silver
has 47 electrons per atom, and its molar mass is 107.87 g/mol.
N= NAm =
M
(6.02 x 1023 atoms/mol x (10 g)
107.87g/mol
N=5.4 x 1022 atom
q= NZe=5.4 x 1022 x 47=25x1023 e
________________________________________________________________________________
me=9.109 ×10−31 kg
F
qq
1 q1q2
 K 1 22 ,
2
4o r
r
mp=1.67×10−27 kg
K
1
1 qc
E (r ) 
4o
4o r 2
e=1.6x10-19 C NA=6.023x1023 mole-1
F x = q t E x V (r )  K
q
r
U tc = qt Vc
Faculty of dentistry
Department of Basic Sciences
Date: 9-4-2011
Quiz 1
Time: 1/2h
Student name …………………………
ID...........................................
Circle the correct answer:
1-Two positive point charges Q and 2Q are separated by a distance R. If the charge Q experiences a
force of magnitude F when the separation is R, what is the magnitude of the force on the charge 2Q
when the separation is 2R ?
(A) F/4
(B) F
(C) 4F
(D) F/2
(E) 2F
2. Choose the correct statement concerning electric field lines:
A. field lines may cross
B. field lines are close together where the field is large
C. field lines point away from a negatively charged particle
D. a charged point particle released from rest moves along a field line
E. none of these are correct
3-An electron is accelerated from rest through a potential difference V . Its final speed is proportional to:
A. V
C. √ V
B. V 2
E. 1/√ V
D. 1/V
The rate at which electrical energy is used may be measured in:
A. watt/second
B. watt·second
C. watt
D. joule·second
E. kilowatt·hour
Current is a measure of:
A. force that moves a charge past a point
B. resistance to the movement of a charge past a point
C. energy used to move a charge past a point
D. amount of charge that moves past a point per unit time
E. speed with which a charge moves past a point
Answer one problem only:
1- The potential difference between the accelerating plates of a TV set is about 25 kV. If the distance
between the plates is 1.5 cm, find the magnitude of the uniform electric field in the region between the plates.
V=25 x 103 V
d= 1.5 x 1-2 m
V=Ed
E=V/d
E=(25 x 103 ) /(1.5 x 1-2 )= 16.6 x105 V/m
2- Two identical conducting spheres, A and B, carry equal amounts of charge(q) and are fixed a distance apart
large compared with their diameters. They repel each other with an electrical force of 220  N. Suppose now
that a third identical sphere C, having an insulating handle and initially uncharged, is touched first to sphere A,
then to sphere B, and finally removed. Find the magnitude force between spheres A and B now.
In the 1st case
F1=k (qAqB)/r2= k (q2)/r2 = 220 x 10 -6 N
In the 2nd case
F2=k (qAqB)/r2= (1/2q x 3/4q)/r2 = 3/8( k (q2)/r2 )
Substituting the value of k (q2)/r2 =220 x 10 -6 N
Then F2 =3/8 x 220 x 10 -6 = 82.5 x 10 -6 N = 82.5 N
_________________________________________________________________________________
me=9.109 ×10−31 kg
E (r ) 
qc
4o r 2
1
mp=1.67×10−27 kg
F x = q t E x V (r )  K
q
r
U tc = qt Vc
e=1.6x10-19 C
NA=6.023x1023 mole-1