IB Physics Quiz: Apply Electric And Magnetic Fields

20 questions · exam conditions

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Apply Electric And Magnetic FieldsQuestion 1 of 20

An electron with initial velocity $v_0$ enters a region of uniform electric field $E$ that is directed opposite to its velocity. The electron comes to rest after travelling a distance $d$ . If the initial velocity is doubled to $2v_0$ , what new distance will the electron travel before coming to rest?

d

\sqrt{2}d

2d

4d

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IB Physics Quiz

IB Physics Quiz: Apply Electric And Magnetic Fields

Practice Apply Electric And Magnetic Fields in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Apply Electric And Magnetic Fields, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$d$
$\sqrt{2}d$
$2d$
$4d$ (correct answer)

Explanation: The force on the electron is constant (

F=eE

), so its acceleration

a

is also constant (

a=eE/m_e

). Using the kinematic equation

v^2 = u^2 + 2as

, with final velocity

v=0

, initial velocity

u=v_0

, and displacement

s=d

, we have

0 = v_0^2 - 2ad

. This gives

d = \frac{v_0^2}{2a}

. The stopping distance

d

is proportional to the square of the initial velocity (

d \propto v_0^2

). If the initial velocity is doubled, the stopping distance will be multiplied by a factor of

2^2 = 4

Question 2

Two identical small conducting spheres carry charges of $-2.0 \times 10^{-9}$ C and $+6.0 \times 10^{-9}$ C. They are separated by a distance $r$ and exert an attractive force of magnitude $F_1$ on each other. The spheres are brought into contact and then returned to the same separation $r$ . What is the new force $F_2$ between them?

$F_1/3$ , repulsive (correct answer)
$F_1/4$ , repulsive
$4F_1/3$ , repulsive
$3F_1/4$ , attractive

Explanation: Initially, the magnitude of the force is

F_1 = k \frac{|(-2\times 10^{-9})(6\times 10^{-9})|}{r^2} = 12 \times 10^{-18} \frac{k}{r^2}

. When the identical spheres touch, charge is conserved and shared equally. The total charge is

-2.0 + 6.0 = +4.0 \times 10^{-9}

C. Each sphere will have a charge of

q' = +2.0 \times 10^{-9}

C. The new force is

F_2 = k \frac{(2\times 10^{-9})^2}{r^2} = 4 \times 10^{-18} \frac{k}{r^2}

. The ratio of the magnitudes is

\frac{F_2}{F_1} = \frac{4}{12} = \frac{1}{3}

. Since the final charges have the same sign, the new force is repulsive.

Question 3

An experiment measures the static charge on several different objects. Which of the following measured charge values is most likely the result of experimental error? (The elementary charge is $e = 1.60 \times 10^{-19}$ C)

$+4.80 \times 10^{-19}$ C
$-8.00 \times 10^{-19}$ C
$+2.40 \times 10^{-19}$ C (correct answer)
$-3.20 \times 10^{-19}$ C

Explanation: The principle of charge quantization states that any observable charge must be an integer multiple of the elementary charge

e

. To check this, we divide each measured value by

e

. A:

(4.80 \times 10^{-19}) / (1.60 \times 10^{-19}) = 3

. B:

(-8.00 \times 10^{-19}) / (1.60 \times 10^{-19}) = -5

. D:

(-3.20 \times 10^{-19}) / (1.60 \times 10^{-19}) = -2

. These are all integers. C:

(2.40 \times 10^{-19}) / (1.60 \times 10^{-19}) = 1.5

. Since this is not an integer, this charge value is not possible and indicates an error.

Question 4

A small positive test charge $+q_0$ is placed at a point P in an electric field where the field strength is $E$ . The force on the test charge is $F_0$ . The charge $+q_0$ is then replaced by a charge $-2q_0$ . What are the new force on this charge and the electric field strength at point P?

Force is $-2F_0$ ; Field is $E$ . (correct answer)
Force is $-2F_0$ ; Field is $-2E$ .
Force is $F_0/2$ ; Field is $E$ .
Force is $2F_0$ ; Field is $E$ .

Explanation: The electric field

E

at a point is defined by the source charges and is independent of any test charge placed there. Therefore, the field at P remains

E

. The force on a charge is given by

F = qE

. Initially,

F_0 = q_0E

. For the new charge

q' = -2q_0

, the new force is

F' = (-2q_0)E = -2(q_0E) = -2F_0

. The force is doubled in magnitude and reversed in direction.

Question 5

The magnitude of the electrostatic force between a proton and an electron is $F_E$ , and the magnitude of the gravitational force between them is $F_G$ . What is the order of magnitude of the ratio $F_E / F_G$ ?

$10^{20}$
$10^{29}$
$10^{36}$
$10^{39}$ (correct answer)

Explanation: The ratio is given by

\frac{F_E}{F_G} = \frac{k e^2 / r^2}{G m_p m_e / r^2} = \frac{k e^2}{G m_p m_e}

. Using approximate values from the data booklet:

k \approx 9 \times 10^9

e \approx 1.6 \times 10^{-19}

G \approx 6.7 \times 10^{-11}

m_p \approx 1.7 \times 10^{-27}

m_e \approx 9.1 \times 10^{-31}

. The calculation is

\frac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{(6.7 \times 10^{-11})(1.7 \times 10^{-27})(9.1 \times 10^{-31})} \approx \frac{2.3 \times 10^{-28}}{1.0 \times 10^{-67}} \approx 2.3 \times 10^{39}

. The order of magnitude is

10^{39}

Question 6

An electric field pattern is described for a system of two point charges. The field lines are observed to originate from charge $q_1$ and terminate on charge $q_2$ . The number of field lines originating from $q_1$ is twice the number terminating on $q_2$ . What can be deduced about the charges?

$q_1 = +2Q$ and $q_2 = -Q$ (correct answer)
$q_1 = +Q$ and $q_2 = -2Q$
$q_1 = -2Q$ and $q_2 = +Q$
$q_1 = -Q$ and $q_2 = +2Q$

Explanation: Electric field lines originate on positive charges and terminate on negative charges. Since lines originate from

q_1

, it must be positive. Since they terminate on

q_2

, it must be negative. The number of field lines is proportional to the magnitude of the charge. If the number of lines from

q_1

is twice the number ending on

q_2

, then

|q_1| = 2|q_2|

. Combining these deductions,

q_1

is positive,

q_2

is negative, and the magnitude of

q_1

is twice that of

q_2

. This corresponds to a relationship like

q_1 = +2Q

and

q_2 = -Q

Question 7

A point charge $+Q$ is placed at the center of a thick, uncharged, spherical conducting shell. What are the charges $q_{inner}$ and $q_{outer}$ induced on the inner and outer surfaces of the shell, respectively?

$q_{inner} = 0$ , $q_{outer} = 0$
$q_{inner} = -Q$ , $q_{outer} = +Q$ (correct answer)
$q_{inner} = -Q$ , $q_{outer} = 0$
$q_{inner} = +Q$ , $q_{outer} = -Q$

Explanation: The electric field inside the conducting material must be zero. To cancel the field produced by the central charge

+Q

, a charge of

-Q

is induced on the inner surface of the shell. Since the shell was initially uncharged, by conservation of charge, a net charge of

+Q

must appear on the outer surface to maintain overall neutrality of the shell. Thus,

q_{inner} = -Q

and

q_{outer} = +Q

Question 8

A charge $+q$ is moved from point X to point Y in the electric field of a fixed charge $+Q$ . Point X is at a distance $3R$ from $+Q$ , and point Y is at a distance $R$ from $+Q$ . What is the work done by the electric field on the charge $+q$ during this process?

$+\frac{2kQq}{3R}$
$-\frac{2kQq}{3R}$ (correct answer)
$+\frac{8kQq}{9R^2}$
$-\frac{8kQq}{9R^2}$

Explanation: Work done by the electric field is the negative of the change in electric potential energy,

W_{field} = -\Delta E_p = -(E_{p,final} - E_{p,initial})

. The electric potential energy is

E_p = \frac{kQq}{r}

E_{p,initial} = \frac{kQq}{3R}

and

E_{p,final} = \frac{kQq}{R}

. Thus,

W_{field} = -(\frac{kQq}{R} - \frac{kQq}{3R}) = -kQq(\frac{3-1}{3R}) = -\frac{2kQq}{3R}

. The work is negative because the repulsive field exerts a force opposite to the direction of motion.

Question 9

An alpha particle (charge +2e, mass ≈ 4u) and a proton (charge +e, mass ≈ 1u) enter a uniform electric field at the same velocity, perpendicular to the field lines. Which statement correctly compares their subsequent motion?

The alpha particle is deflected more because it has a greater charge.
The proton is deflected more because it has a smaller charge-to-mass ratio.
The proton is deflected more because it has a greater charge-to-mass ratio. (correct answer)
Both particles are deflected by the same amount because the field is uniform.

Explanation: The path is parabolic. The transverse deflection depends on the acceleration perpendicular to the initial velocity, given by

a = F/m = qE/m

. The particle with the larger charge-to-mass ratio (q/m) will experience greater acceleration and thus greater deflection. For the proton:

q/m \approx e/u

. For the alpha particle:

q/m \approx 2e/4u = (1/2)(e/u)

. The proton has a charge-to-mass ratio twice that of the alpha particle, so it will be deflected more.

Question 10

Three point charges are placed at the vertices of an equilateral triangle of side length $L$ . Two of the charges are $+Q$ , and the third is $-Q$ . What is the magnitude of the net electric force on one of the $+Q$ charges?

$\frac{kQ^2}{L^2}$
$\sqrt{2} \frac{kQ^2}{L^2}$
$\sqrt{3} \frac{kQ^2}{L^2}$ (correct answer)
$2 \frac{kQ^2}{L^2}$

Explanation: Consider one of the

+Q

charges. It experiences a repulsive force from the other

+Q

charge and an attractive force from the

-Q

charge. Both forces have the same magnitude,

F = \frac{kQ^2}{L^2}

. The angle between the sides of an equilateral triangle is 60°. The repulsive force acts along the line connecting the two

+Q

charges, and the attractive force acts along the line connecting the

+Q

and

-Q

charges. The angle between these two force vectors is 60°. The magnitude of the resultant vector is found using the law of cosines:

F_{net}^2 = F^2 + F^2 + 2F^2 \cos(60^{\circ}) = 2F^2 + 2F^2(1/2) = 3F^2

. Therefore,

F_{net} = \sqrt{3}F = \sqrt{3} \frac{kQ^2}{L^2}

Question 11

In an experiment, an oil drop of mass $m$ and charge $q$ is held stationary by a uniform electric field $E$ between two horizontal plates. The gravitational field strength is $g$ . If the magnitude of the charge on the drop spontaneously halves while the electric field remains constant, what is the initial acceleration of the drop?

$g/2$ downwards (correct answer)
$g/2$ upwards
$g$ downwards
$2g$ downwards

Explanation: Initially, the drop is stationary, so the net force is zero. The upward electric force

F_E = qE

balances the downward gravitational force

F_g = mg

. So,

qE = mg

. When the charge becomes

q/2

, the new electric force is

F'_E = (q/2)E = (mg)/2

. The gravitational force remains

mg

. The net force is now

F_{net} = F_g - F'_E = mg - mg/2 = mg/2

downwards. Using Newton's second law,

F_{net} = ma'

, we get

ma' = mg/2

, which means the new acceleration is

a' = g/2

downwards.

Question 12

Two point charges are separated by a distance $d$ , and the magnitude of the force between them is $F$ . If the distance is increased to $3d$ and the magnitude of each charge is doubled, what is the new force magnitude?

$F/9$
$2F/9$
$4F/9$ (correct answer)
$2F/3$

Explanation: Coulomb's Law states

F = k \frac{|q_1 q_2|}{d^2}

. The new charges are

2q_1

and

2q_2

, and the new distance is

3d

. The new force

F'

F' = k \frac{|(2q_1)(2q_2)|}{(3d)^2} = k \frac{4|q_1 q_2|}{9d^2} = \frac{4}{9} \left( k \frac{|q_1 q_2|}{d^2} \right) = \frac{4}{9}F

. The force is multiplied by a factor of 4 due to the charges and divided by a factor of 9 due to the distance.

Question 13

The electric potential at a distance $r$ from a point charge $Q$ is $V$ . The magnitude of the electric field at the same point is $E$ . What is the electric potential at a distance $2r$ from a point charge $2Q$ ?

$V/2$
$V$ (correct answer)
$2V$
$4V$

Explanation: Electric potential from a point charge is given by the formula

V = \frac{kQ}{r}

. The information about the electric field

E

is extraneous. The new potential

V'

is for a charge

Q' = 2Q

at a distance

r' = 2r

. Substituting these into the formula gives

V' = \frac{k(2Q)}{(2r)} = \frac{kQ}{r} = V

. The potential remains the same.

Question 14

A charge of +Q is placed at one corner of a square, and a charge of -Q is placed at the diagonally opposite corner. The side length of the square is $L$ . What is the magnitude of the net electric field at one of the other corners?

$\frac{kQ}{L^2}$
$\sqrt{2} \frac{kQ}{L^2}$ (correct answer)
$\sqrt{5} \frac{kQ}{L^2}$
$2 \frac{kQ}{L^2}$

Explanation: Let the corner in question be Q. The charge +Q is at a distance

L

from Q, and the charge -Q is also at a distance

L

from Q. The electric field from +Q at Q,

E_1

, has magnitude

kQ/L^2

and points away from +Q. The electric field from -Q at Q,

E_2

, has magnitude

kQ/L^2

and points towards -Q. These two field vectors are perpendicular to each other. The magnitude of the net electric field is found using the Pythagorean theorem:

E_{net} = \sqrt{E_1^2 + E_2^2} = \sqrt{(\frac{kQ}{L^2})^2 + (\frac{kQ}{L^2})^2} = \sqrt{2(\frac{kQ}{L^2})^2} = \sqrt{2} \frac{kQ}{L^2}

Question 15

Two point charges, $+2q$ and $-q$ , are fixed a distance $d$ apart. Point P is located on the line connecting the charges, but at a location external to them, a distance $d$ from the $-q$ charge and $2d$ from the $+2q$ charge. What is the electric potential at point P? (Assume potential is zero at infinity.)

$0$ (correct answer)
$+\frac{kq}{2d}$
$-\frac{kq}{d}$
$+\frac{3kq}{2d}$

Explanation: Electric potential is a scalar quantity, so the total potential at a point is the algebraic sum of the potentials due to each charge. The potential from the

+2q

charge is

V_1 = \frac{k(+2q)}{2d} = \frac{kq}{d}

. The potential from the

-q

charge is

V_2 = \frac{k(-q)}{d} = -\frac{kq}{d}

. The total potential at point P is

V_P = V_1 + V_2 = \frac{kq}{d} - \frac{kq}{d} = 0

Question 16

A solid, uncharged conducting sphere is placed in a uniform external electric field that points to the right. Which statement best describes the net electric field inside the material of the sphere after electrostatic equilibrium is reached?

The net field is zero everywhere inside the sphere. (correct answer)
The net field is uniform and has the same magnitude as the external field.
The net field is non-uniform and points to the left.
The net field points to the right, but its magnitude is less than the external field.

Explanation: In electrostatic equilibrium, the free charges within a conductor rearrange themselves to cancel any external electric field. Electrons migrate to the surface on the side opposing the field direction, creating an induced internal electric field that is equal in magnitude and opposite in direction to the external field. The vector sum of the external field and the induced field is zero everywhere inside the conductor.

Question 17

A small object with mass $m$ and positive charge $q$ is in a vacuum where there is a uniform gravitational field $g$ acting downwards and a uniform electric field $E$ acting horizontally. The object is released from rest. What is the shape of its subsequent path?

A parabola opening downwards.
A straight line directed horizontally.
A parabola opening diagonally.
A straight line directed diagonally downwards. (correct answer)

Explanation: The object is subject to two constant forces: a gravitational force

F_g = mg

downwards and an electric force

F_E = qE

horizontally. The net force,

\vec{F}_{net} = \vec{F}_g + \vec{F}_E

, is the vector sum of these two forces. Since both forces are constant in magnitude and direction, the net force is also constant in magnitude and direction (diagonally downwards). By Newton's second law, the acceleration

\vec{a} = \vec{F}_{net}/m

is also constant. An object starting from rest and moving with constant acceleration follows a straight line path in the direction of the acceleration.

Question 18

A proton moves in a helical path through a region containing both uniform electric and magnetic fields. The electric field $\vec{E}$ is parallel to the magnetic field $\vec{B}$ , and both point in the positive z-direction. Which statement best describes the motion of the proton?

Circular motion in the xy-plane with constant radius, while accelerating in the z-direction due to the electric field (correct answer)
Circular motion in the xy-plane with increasing radius, as the electric field increases the proton's total kinetic energy
Linear motion in the z-direction with constant velocity, while the magnetic field causes oscillation in the xy-plane
Elliptical motion in the xy-plane with decreasing radius, as the electric field opposes the magnetic force

Explanation: The magnetic force

\vec{F} = q\vec{v} \times \vec{B}

is always perpendicular to both velocity and magnetic field, so it only affects motion perpendicular to

\vec{B}

(in the xy-plane). This creates circular motion with radius

r = mv_{\perp}/(qB)

. Since the magnetic force does no work,

v_{\perp}

remains constant, so the radius stays constant. The electric force

q\vec{E}

acts parallel to

\vec{B}

(z-direction), causing acceleration in the z-direction. B is wrong because

v_{\perp}

doesn't change, so radius is constant. C is wrong because motion in z-direction is accelerated, not constant velocity. D is wrong because there's no mechanism for decreasing radius, and the electric field doesn't oppose magnetic force.

Question 19

Two identical positive point charges are fixed at positions $(-a, 0)$ and $(+a, 0)$ on the x-axis. A third positive point charge is placed at position $(0, b)$ on the y-axis, where $b < a$ . If the third charge is displaced slightly in the positive y-direction, what type of motion will result?

Simple harmonic motion toward the equilibrium position, since the net force is restoring and proportional to displacement
Circular motion around the y-axis, due to the symmetric arrangement of the charges creating a central force
Accelerated motion away from the x-axis, since the equilibrium at $(0, b)$ is unstable in the y-direction (correct answer)
Oscillatory motion with increasing amplitude, as the restoring force decreases with distance from equilibrium

Explanation: When analyzing equilibrium in electrostatic systems, you need to determine whether a displaced charge experiences forces that restore it to equilibrium (stable) or push it further away (unstable). The key is examining how the net force changes with small displacements. At position

(0, b)

, the third charge is in equilibrium because the horizontal components of the repulsive forces from the two fixed charges cancel out, while the vertical components balance. However, when you displace this charge slightly upward to

(0, b + \delta y)

, the geometry changes crucially. As the third charge moves farther from the x-axis, the angles between the force vectors and the y-axis decrease. This means the vertical components of the repulsive forces from both fixed charges become larger, creating a net upward force that grows with displacement. Since the force pushes the charge further from equilibrium rather than back toward it, the equilibrium is unstable, leading to accelerated motion away from the x-axis. Option A is wrong because the force isn't restoring—it pushes away from equilibrium. Option B incorrectly assumes circular motion; the forces are purely repulsive with no mechanism for circular orbits. Option D describes unstable oscillation, but once displaced, the charge won't return to oscillate—it will simply accelerate away. Remember: In electrostatic equilibrium problems, always check stability by examining whether small displacements result in restoring forces (stable) or forces that increase displacement (unstable). The mathematical test involves checking the sign of the force derivative at equilibrium.

Question 20

A circular loop of wire with radius $R$ carries current $I$ in a clockwise direction when viewed from above. A small compass is placed at various positions around the loop. At which position will the compass needle point in the direction most nearly perpendicular to the plane of the loop?

At the center of the loop, where the magnetic field is strongest and most uniform (correct answer)
At a point on the axis of the loop, far from the center compared to the radius
At a point in the plane of the loop, outside the wire at distance $R$ from the center
At a point in the plane of the loop, inside the wire at distance $R/2$ from the center

Explanation: Using the right-hand rule, a clockwise current (viewed from above) produces a magnetic field pointing downward (into the page) at the center of the loop. This field direction is perpendicular to the plane of the loop. At the center, the field is purely perpendicular with no component in the plane. B is incorrect because on the axis far from the center, the field is still perpendicular but weaker, and 'most nearly' suggests we want the strongest perpendicular field. C and D are incorrect because at points in the plane of the loop, the magnetic field has components parallel to the plane, so the field direction is not perpendicular to the plane.