Sec Phys Electrostatics Problems
Sec Phys Electrostatics Problems
Sec Phys Electrostatics Problems
C UULLTT Y O
OFF EE
DD U AT
UC C AI O
T INO N
Department of
Curriculum and Pedagogy
Physics
Electrostatics Problems
Science and Mathematics
Education Research Group
Answer: A
Justification: Let us go through the operation step-by-step:
A B
Initially, both spheres A and B have no charges.
Once these spheres are touching each other
they become one conducting object.
A B When a positively charged rod comes close to
sphere A without touching it, sphere A becomes
negatively charged by induction. This means that
the negative charges in both spheres A and B will
A B
move towards the positively charged rod (opposite
Electrons move
charges attract).
from B to A Therefore sphere B becomes positively charged
since most of the electrons will have gone to the
side of sphere A.
Questioncontinued
Solution Title
Attraction
Repulsion
1 2 2 3
Answer: E
Justification: By looking at the diagram, we can see that
balls 1 and 2 attract each other and balls 2 and 3 repel each
other.
We know that like charges repel each other and opposite
charges attract each other. However, we were also told that
the balls could also have no charge (i.e. neutral charge).
If an object with a neutral charge is brought close to a
charged object, they will always attract each other. This is
due to induction – the charged object will induce the
electrons within the neutral object to move from one side of
the object to the other – see next page for diagram.
Questioncontinued
Solution Title
Below are two examples of induction, one with a positively charged
object, and another with a negatively charged object. In each case, it is
the electrons which move (they are more loosely bound than the
protons):
Neutral object Neutral object
ATTRACTION ATTRACTION
1 2 3
Electrostatics Problems III
Question
continued
Title
A. 1
B. 3
C. 2 and 3
D. 2 and 4
E. 2, 3, and 4
Question Title
Solution
Answer: B
Justification: We need to look at each of the positions that the
negative point charge could take:
1) In this case the repulsive force between the negative
charge and the negative point charge will dominate the weak
attractive force exerted by the positive charge on the
negative point charge. Thus, the negative point charge will
move to the left.
2) In this case the repulsive force and the attractive force
exerted by the 2 big charges will move the negative point
charge to the right no matter where you put the point charge
between the 2 big charges.
Questioncontinued
Solution Title
REPULSION
ATTRACTION
-
Negative
point
charge
Questioncontinued
Solution Title 2
4) In this case the charge will move to the right due to the
repulsion by the negative charge and the attraction of the positive
charge.
A. q1 = 3q
B. q2 = – 3q
C. The magnitude of F2on1 is also tripled and its direction is changed.
D. The magnitude of F2on1 is also tripled.
E. The magnitude of F2on1 is one third of its original value.
Question Title
Solution
Answer: D
Justification: Since the force exerted by each charge is always
equal in magnitude, this means that when F1on2 is tripled, so is
F2on1. If we look at all the possible answers we see that this is the
only answer that must be true:
A) This would cause both F1on2 and F2on1 to triple, but it does not
have to be true because there are other ways to get these
forces to triple in magnitude.
B) Same as A - this would cause both F1on2 and F2on1 to triple, but
it does not have to be true because there are other ways to
get these forces to triple in magnitude.
C) Although the change in magnitude is correct, the change in
direction is not necessary.
Questioncontinued
Solution Title
D. This must be true if F1on2 is tripled.
E. This is incorrect because F1on2 and F2on1 must be equal
A. i and iii
B. ii only
C. iii
D. i, iv and v
E. i and iv
Question Title
Solution
Answer: D
Justification: To begin, the original force exerted by each charge is:
Next, we want to see what changes in the variables will give a new
force, F2, which is double the magnitude of F. In other words:
F2 = 2F
Questioncontinued
Solution Title
From this we see that (i), (iv) and (v) give the answer F2 = 2F, therefore
the answer is D.
Question TitleProblems VI
Electrostatics
Two charges, 1 and 2, with charges +q and –q respectively are placed a
distance r apart as shown in the diagram below. According to Coulomb’s
Law the magnitude of the force exerted on each charge by the other is:
Electrostatics Problems VI
Question
continued
Title
Question Title
Solution
Questioncontinued
Solution Title
Question TitleProblems VII
Electrostatics
Answer: E
Justification: We know that the magnitude of the electric force
between the charges can be represented as:
This electric force is equal to the net force of the charges (we are
ignoring gravity). Each of the charges experiences an acceleration due
to this force, and since the charges have equal mass (m), this can be
represented as:
Since the force Fnew has increased to 4FE, and the acceleration is
proportional to the force, we know that the new acceleration anew must
also have increased four-fold: anew = 4a
A. 10 N/C
B. 20 N/C
C. 40 N/C
D. 80 N/C
Question Title
Solution
Answer: B
Justification: Let the electric field strength be denoted by 𝐸. The
magnitude of the electric field strength (𝐸) is defined as the force (𝐹)
𝐹
per charge (𝑞) on the source charge (𝑄). In other words, 𝐸 = , where
𝑞
𝑘𝑞𝑄
𝐹= is the electric force given by Coulomb's law, k is the
𝑑2
𝑚2
Coulomb's law constant (𝑘 = 9.0 × 109 𝑁 ), and d is the distance
𝐶2
between the centers of 𝑞 and 𝑄.
𝑘𝑞𝑄
So we need to use the expression, 𝐸 = . Simplifying this
𝑞𝑑 2
𝑘𝑄
expression gives, 𝐸 = .
𝑑2
Questioncontinued
Solution Title
Answer: B
In our case, since 𝑄 and 𝑑 are both doubled, the new field strength is
𝑘 (2𝑄) 2 𝑘𝑄 1
𝐸𝑛𝑒𝑤 = 2 , which can be simplified to get 𝐸𝑛𝑒𝑤 = × 2 = 𝐸.
(2𝑑) 4 𝑑 2
1 1
Thus, the new field strength is 𝐸𝑛𝑒𝑤 = 𝐸 = × 40 𝑁/𝐶 = 20𝑁/𝐶.
2 2
Two point charges (C1 and C2) are fixed as shown in the setup
below. Now consider a third negative charge C3 with charge -q that
you can place anywhere you want in regions A, B, C, or D. In which
region could you place charge C3 so that the net force on it is zero?
C1 C2 A. Region A
B. Region B
C. Region C
D. Region D
Question Title
Solution
d/3
D.
q q
Question Title
Solution
Answer: B
Justification: Recall that electric potential energy of a system of
electric charges 𝐸𝑃 depends on two quantities: 1) electric
charges involved and 2) the distances between them.
3𝑞 × 3𝑞 𝑞2
For system B: 𝐸𝑃 = 𝑘 = 9𝑘
𝑑 𝑑
2𝑞 ×10𝑞 𝑞2
For system C: 𝐸𝑃 = 𝑘 = 10𝑘
2𝑑 𝑑
𝑞 ×𝑞 𝑞2
For system D: 𝐸𝑃 = 𝑘 = 3𝑘
𝑑/3 𝑑
𝑞2
Since 𝑘 is common to all of the above expressions, we note that the
𝑑
numerical coefficients determine the rank of the electric potential
energies (i.e., 10 > 9 > 4 > 3). Thus B is the correct answer.
Question TitleProblems XI
Electrostatics
In each of the four scenarios listed below, the two charges remain
fixed in place as shown. Rank the forces acting between the two
charges from the greatest to the least.
d A. C >B >A>D
A.
4q q B. C>B=D>A
C. B=D>C>A
d
B. D. B +D >A>C
3q 3q
E. A>C>B=D
2d
C.
2q 10q
d/3
D.
q q
Question Title
Solution
Answer: C
Justification: Recall that the electric force is a fundamental force of
the universe that exists between all charged particles. For example,
the electric force is responsible for chemical bonds. The strength of
the electric force between any two charged objects depends on the
amount of charge that each object contains and also on the distance
between the two charges. From Coulomb's law, we know that the
𝑞 𝑞
electric force is given by 𝐹 = 𝑘 1 2 2 , where 𝑘 is the Coulomb's law
𝑟
constant, 𝑞1 and 𝑞2 are point charges, and 𝑟 is the distance between
the two point charges.
Note that 𝐹 is proportional to the amount of charge and also inversely
proportional to the square of the distance between the charges.
Questioncontinued
Solution Title
Answer: C
4𝑞 × 𝑞 𝑞2
For system A: 𝐹 = 𝑘 = 4𝑘 2
𝑑2 𝑑
3𝑞 × 3𝑞 𝑞2
For system B: 𝐹 = 𝑘 = 9𝑘 2
𝑑 𝑑
2𝑞 ×10𝑞 𝑞2
For system C: 𝐹 = 𝑘 = 5𝑘 2
(2𝑑)2 𝑑
𝑞 ×𝑞 𝑞2
For system D: 𝐹 = 𝑘 = 9𝑘 2
(𝑑/3)2 𝑑
𝑞2
Since 𝑘 2 is common to all of the above expressions, we note that
𝑑
the numerical coefficients determine the rank of the electric forces
(i.e. 9 = 9 > 5 > 4). Thus C is the correct answer.
Question TitleProblems XII
Electrostatics
Given the following electric field diagrams:
Answer: C FIX IT
Justification: Recall that the direction is defined as the direction that
a positive test charge would be pushed when placed in the electric
field. The electric field direction of a positively charged object is
always directed away from the object. And also, the electric field
direction of a negatively charged object is directed towards the object.
Questioncontinued
Solution Title
Answer: C
Since the field direction is directed away from (a) but towards (b) and
(c), we know that the relative charges of (a,b,c) = (+,-,-)
Note that the field lines allow us to not only visualize the direction of
the electric field, but also to qualitatively get the magnitude of the field
through the density of the field lines. From (a), (b), and (c), we can
see that the density of the electric field lines in (c) is twice that of (a)
or (b). We would expect the magnitude of the charge in (c) to also be
twice as strong as (a) or (b). Thus, the answer choice C is correct.
Question TitleProblems XIII
Electrostatics
A. B>D >A>C
B. B>D>C>A
C. D>B>C>A
D. D >B >A>C
E. A>B>D>C
Question Title
Solution
Answer: D
Justification: We need to understand the concept of the electric field
being zero inside of a closed conducting surface of an object, which
was demonstrated by Michael Faraday in the 19th century. Suppose
to the contrary, if an electric field were to exist below the surface of
the conductor, then the electric field would exert a force on electrons
present there. This implies that electrons would be in motion.
However, the assumption that we made was that for objects at
electrostatic equilibrium, charged particles are not in motion. So if
charged particles are in motion, then the object is not in electrostatic
equilibrium. Thus, if we assume that the conductor is at electrostatic
equilibrium, then the net force on the electrons within the conductor is
zero. So at point C, the electric field is zero.
Questioncontinued
Solution Title
Answer: D
For conductors at electrostatic equilibrium, the electric fields are
strongest at regions along the surface where the object is most
curved. The curvature of the surface can range from flat regions to
that of being a blunt point, as shown below.