Republic of the Philippines
Pamantasan ng Cabuyao
(UNIVERSITY
OF CABUYAO)
Katapatan Subd. Brgy. Banay-banay, City of Cabuyao, Laguna
COLLEGE OF ENGINEERING
ELLEZER JAY A. BALANE
2ECE-B
PHYSICS 2
JUANCHO ARAGON
I. Multiple Choices
Only write the letter of the correct answer.
1. What is an electric field? B
a) The flow of electric charge
b) A region in space where electric forces can be observed
c) The movement of electrons in a conductor
d) The resistance of a material to electric current
2. Which of the following units is used to measure electric field strength? Newton per
Coulomb ( none of the choices )
a) Volts
b) b) Amps
c) Ohms
d) Coulombs per Newton
3. What is the direction of the electric field lines around a positive point charge? B
a) Radially inward
b) Radially outward
c) Random
d) There are no electric field lines around a positive point charge.
a) If you double the distance between a point charge and a test charge, how does the electric
field strength at the new distance compare to the original distance? It become one-fourth as
strong ( none of the choices )
b) It becomes four times stronger.
c) It becomes half as strong.
d) It remains the same.
e) It becomes twice as strong.
Republic of the Philippines
Pamantasan ng Cabuyao
(UNIVERSITY
OF CABUYAO)
Katapatan Subd. Brgy. Banay-banay, City of Cabuyao, Laguna
COLLEGE OF ENGINEERING
4. What is the formula for calculating the electric field (E) produced by a point charge (Q)
at a distance (r) from the charge? A
a) E = Q / (4πε₀r²)
b) E = Q / (2πε₀r)
c) E = Q / (ε₀r)
d) E = Q / (πε₀r)
5. Which fundamental constant of nature is associated with the electric field? D
a) Speed of light (c)
b) Planck's constant (h)
c) Gravitational constant (G)
d) Permittivity of free space (ε₀)
6. True or False: Electric field lines always point from positive charges to negative charges.
True
7. What happens to the magnitude of the electric field when you increase the magnitude of
the source charge in the formula E = kQ / r² (where k is the electrostatic constant)? A
a) The electric field increases.
b) The electric field decreases.
c) The electric field remains the same.
d) The electric field depends on other factors, not just the source charge.
8. What is the relationship between the electric field strength (E) and the distance (r) from a
point charge in a vacuum, according to Coulomb's law? B
a) E is directly proportional to r.
b) E is inversely proportional to r.
c) E is directly proportional to the square of r.
d) E is independent of r.
9. Two point charges, +Q and -Q, are placed a certain distance apart. What can you say
about the direction of the electric field at the midpoint between these two charges? D
a) The electric field points away from the +Q charge.
Republic of the Philippines
Pamantasan ng Cabuyao
(UNIVERSITY
OF CABUYAO)
Katapatan Subd. Brgy. Banay-banay, City of Cabuyao, Laguna
COLLEGE OF ENGINEERING
b) The electric field points away from the -Q charge.
c) The electric field points toward the midpoint between the charges.
d) The electric field cancels out to zero at the midpoint.
I. Problem Solving: Show your solution.
(2 points) A solid ball with 0.5 meters radius has 10 μC electric charge in its center. Determine the
electrical flux pass through the solid ball.
Solution:
r = 0.5 m
Q = 10 × 10−6 C
Nm2
𝑘 = 9 × 10
C2
9
𝛷 =?
𝛷 = 𝐸𝐴
𝑘𝑞
)(4𝜋𝑟 2 )
𝑟2
𝑘𝑞
𝛷 = ( 2 )(4𝜋𝑟 2 )
𝑟
𝛷=(
𝛷 = (𝑘𝑞)(4𝜋)
𝛷 = (9 × 109
Nm2
)(10 × 10−6 C)(4𝜋)
C2
𝛷 = 1,130,973.3552923256
𝛷 = 1.13 × 106
Nm2
C
Nm2
C
(3 points) Two point charges are placed in the xy-plane. Charge Q1 = +3 μC is located at the origin
(0,0), and charge Q2 = -5 μC is located at the point (4 m, 0). Calculate the electric field strength at
a point P located on the positive y-axis, 3 meters above the origin (0, 3 m).
Republic of the Philippines
Pamantasan ng Cabuyao
(UNIVERSITY
OF CABUYAO)
Katapatan Subd. Brgy. Banay-banay, City of Cabuyao, Laguna
COLLEGE OF ENGINEERING
Solution:
r1 = 3 m
r2 = √32 + 42 = 5m
Q1 = +3 × 10−6 C
Q2 = −5 × 10−6 C
𝑘 = 9 × 109
Nm2
C2
𝐸 =?
𝐸1 =
𝑘𝑞
𝑟2
Nm2
)(3 × 10−6 C)
C2
32
(9 × 109
𝐸1 =
Nm2
)(3 × 10−6 C)
C2
(3𝑚)2
(9 × 109
𝐸1 =
𝐸1 = 3,000
𝐸2 =
(9 × 109
𝐸2 =
(9 × 109
𝐸2 =
N
C
𝑘𝑞
𝑟2
Nm2
)(−5 × 10−6 C)
C2
52
Nm2
)(−5 × 10−6 C)
C2 m
(5𝑚)2
𝐸2 = −1,800
N
C
𝐸𝑇 = 𝐸1 + 𝐸2
𝐸𝑇 = 3,000
N
C
N
+ (−1,800 C )
𝐸𝑇 = 1,200
N
C