Apply Electric & Magnetic Fields - IB Physics

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?
Answer: $F_1/3$ , repulsive 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.
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?
Answer: $q_{inner} = -Q$ , $q_{outer} = +Q$ 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 $q_{outer} = +Q$ .
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?
Answer: $g/2$ downwards 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.
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?
Answer: A straight line directed diagonally downwards. 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.
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?
Answer: Circular motion in the xy-plane with constant radius, while accelerating in the z-direction due to the electric field 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.
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.)
Answer: $0$ 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 $V_P = V_1 + V_2 = \frac{kq}{d} - \frac{kq}{d} = 0$ .
A uniform electric field $\vec{E} = E_0 \hat{j}$ exists in a region of space. An electron with initial velocity $\vec{v_0} = v_0 \hat{i}$ enters this region. After time $t$ , the electron's velocity makes an angle $\theta$ with the positive x-axis. What is the relationship between $\theta$ , the field strength $E_0$ , and the time $t$ ?
Answer: $\tan \theta = \frac{eE_0t}{mv_0}$ , since the electron gains velocity component in the y-direction while maintaining constant x-velocity When a charged particle moves through a uniform electric field, you're analyzing projectile-like motion where the field creates constant acceleration in one direction while the particle maintains constant velocity in the perpendicular direction. Here, the electron enters with velocity $\vec{v_0} = v_0 \hat{i}$ (purely horizontal) and experiences electric field $\vec{E} = E_0 \hat{j}$ (purely vertical). The electric force on the electron is $\vec{F} = -e\vec{E} = -eE_0 \hat{j}$ (negative because electrons have negative charge). This creates acceleration $a_y = -\frac{eE_0}{m}$ in the y-direction. After time $t$ : - x-velocity remains constant: $v_x = v_0$ - y-velocity increases: $v_y = a_y t = -\frac{eE_0}{m}t$ The angle $\theta$ with the positive x-axis satisfies $\tan \theta = \frac{v_y}{v_x} = \frac{-eE_0t/m}{v_0} = -\frac{eE_0t}{mv_0}$ . Taking the magnitude, we get $\tan \theta = \frac{eE_0t}{mv_0}$ , which is answer D. Answer A incorrectly inverts this ratio. Answer B uses sine instead of tangent - sine would relate one velocity component to the total velocity magnitude, not the ratio of components. Answer C uses cosine and has the wrong ratio entirely. Study tip: In projectile motion problems with charged particles, always identify which velocity components remain constant versus which ones change due to the field. The tangent of the deflection angle equals the ratio of the changing component to the constant component.
Which statement about electric fields and equipotential surfaces is always correct?
Answer: The electric field is stronger where equipotential surfaces are closer together. The relationship between electric field strength $E$ and the potential difference $\Delta V$ over a distance $\Delta x$ is $E \approx -\Delta V/\Delta x$ . For a fixed potential difference between adjacent equipotential surfaces, a smaller separation $\Delta x$ implies a larger electric field strength. Choice B is incorrect as the path can be curved. Choice C is incorrect; for a point charge, they are concentric spheres. Choice D is incorrect; work is zero only when moving along a single equipotential surface ( $\Delta V = 0$ ), not between different ones.
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 $F_E / F_G$ ?
Answer: $10^{39}$ 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 $10^{39}$ .
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?
Answer: $\sqrt{2} \frac{kQ}{L^2}$ 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: $E_{net} = \sqrt{E_1^2 + E_2^2} = \sqrt{(\frac{kQ}{L^2})^2 + (\frac{kQ}{L^2})^2} = \sqrt{2(\frac{kQ}{L^2})^2} = \sqrt{2} \frac{kQ}{L^2}$ .
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?
Answer: Accelerated motion away from the x-axis, since the equilibrium at $(0, b)$ is unstable in the y-direction 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.
An electron beam is directed horizontally in a cathode ray tube. The beam passes through crossed electric and magnetic fields arranged so that the electric field points vertically upward and the magnetic field points horizontally into the page. Under what conditions will the electron beam continue in a straight horizontal path?
Answer: When $E = vB$ , where $v$ is the electron's horizontal velocity, making the electric and magnetic forces equal in magnitude For straight-line motion, the net force perpendicular to the velocity must be zero. The electric force on an electron is $F_E = eE$ (downward, since electron has negative charge and E points up). The magnetic force is $F_B = evB$ (upward, using right-hand rule for negative charge). For equilibrium: $eE = evB$ , which gives $E = vB$ . A is incorrect because it compares field magnitudes without considering velocity dependence. C incorrectly includes $v^2$ , which doesn't appear in the force equations. D rearranges the correct relationship but expresses it as a condition on B rather than the more fundamental relationship $E = vB$ .
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?
Answer: At the center of the loop, where the magnetic field is strongest and most uniform 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.
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?
Answer: $\sqrt{3} \frac{kQ^2}{L^2}$ 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, $F_{net} = \sqrt{3}F = \sqrt{3} \frac{kQ^2}{L^2}$ .
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 $2Q$ ?
Answer: $V$ 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.
A rectangular conducting loop moves with velocity $\vec{v}$ through a region where a uniform magnetic field $\vec{B}$ exists. The loop is oriented so that its plane is parallel to $\vec{v}$ and perpendicular to $\vec{B}$ . As the loop enters the magnetic field region, which statement correctly describes the induced current and its effects?
Answer: The induced current creates a magnetic field that opposes the change in flux, resulting in a force that opposes the loop's motion When you encounter problems involving conducting loops moving through magnetic fields, you're dealing with electromagnetic induction governed by Faraday's law and Lenz's law. The key insight is that changing magnetic flux through a loop induces an EMF and current. As the loop enters the magnetic field region, the magnetic flux through it increases from zero to $\Phi = BA$ , where $A$ is the area of the loop inside the field. This changing flux induces an EMF by Faraday's law: $\mathcal{E} = -\frac{d\Phi}{dt}$ . The induced current flows in a direction determined by Lenz's law: it creates a magnetic field that opposes the change causing it. Since flux is increasing as the loop enters, the induced current creates a field opposing the external field within the loop. This current-carrying conductor in the external magnetic field experiences a force $\vec{F} = I\vec{L} \times \vec{B}$ that opposes the loop's motion. Choice A is incorrect because while current direction does reverse between entering and exiting, this doesn't mean forces cancel - the opposing force exists throughout each phase. Choice B contradicts Lenz's law by suggesting the induced field enhances rather than opposes the external field. Choice C incorrectly assumes no flux change occurs; flux clearly increases as more loop area enters the field region. The correct answer is D - the induced current opposes flux changes and creates forces opposing motion, demonstrating energy conservation in electromagnetic systems. Study tip: Remember that Lenz's law always creates opposition - induced effects oppose their causes. This principle helps you quickly eliminate choices suggesting enhancement or acceleration.
Two parallel plates separated by distance $d$ have a potential difference $V$ . An electron enters the region between the plates with initial velocity $v_0$ parallel to the plates. After traveling a horizontal distance $L$ between the plates, the electron's vertical displacement is $y$ . If the plate separation is doubled while keeping all other initial conditions the same, what will be the new vertical displacement?
Answer: $\frac{y}{2}$ , because the electric field strength is halved and acceleration is proportionally reduced The electric field between parallel plates is $E = V/d$ . When $d$ doubles, $E$ becomes $V/(2d) = E/2$ . The acceleration is $a = eE/m$ , so the new acceleration is $a/2$ . The time to cross the plates remains $t = L/v_0$ since horizontal motion is unaffected. The vertical displacement is $y = \frac{1}{2}at^2$ , so with half the acceleration, the displacement becomes $\frac{1}{2}(a/2)t^2 = y/2$ . A incorrectly suggests displacement changes by factor of 4. C incorrectly assumes more space means more displacement, ignoring the weaker field. D incorrectly focuses on total energy rather than the kinematics of motion in the field.
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?
Answer: $q_1 = +2Q$ and $q_2 = -Q$ 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 $q_2 = -Q$ .
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?
Answer: $4d$ 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 $2^2 = 4$ .
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?
Answer: The proton is deflected more because it has a greater charge-to-mass ratio. 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.

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