Electric Fields - AP Physics C Electricity and Magnetism

1. A point charge $Q=+4.0\ \mu\text{C}$ sits at the center of a cube of side $0.60\ \text{m}$. Based on the scenario described, using Gauss's Law determine total electric flux through the cube ($\varepsilon_0=8.85\times10^{-12}$).
   Answer: $\Phi_E=4.5\times10^{5}\ \text{N}\cdot\text{m}^2/\text{C}$ This question tests AP Physics C: Electricity and Magnetism skills, specifically applying Gauss's Law to find total flux through closed surfaces. Gauss's Law states that the total electric flux through any closed surface equals Q_enclosed/ε₀, regardless of surface shape or symmetry. In this scenario, a point charge Q = +4.0 μC is at the center of a cube, so all charge is enclosed and Φ_E = Q/ε₀ = (4.0×10⁻⁶)/(8.85×10⁻¹²) = 4.52×10⁵ N·m²/C. Choice A is correct because it gives the right magnitude with proper units - the cube's size and shape don't affect the total flux, only the enclosed charge matters. Choice B is incorrect because it wrongly claims zero flux due to lack of symmetry - Gauss's Law works for any closed surface, symmetric or not. To help students: Emphasize that while symmetry helps in calculating fields, total flux depends only on enclosed charge for any closed surface. Watch for: confusion between needing symmetry to find field versus finding total flux.
2. A thin ring of radius $R$ holds a total charge $+Q$ distributed uniformly. The magnitude of the electric field along the central axis is given by $E(z) = \frac{kQz}{(z^2 + R^2)^{3/2}}$, where $z$ is the distance from the center of the ring. At what distance $z$ from the center is the magnitude of the electric field at its maximum value?
   Answer: $z = R/\sqrt{2}$ To find the maximum field, we must take the derivative of $E(z)$ with respect to $z$ and set it to zero. Using the quotient rule, $dE/dz = kQ \frac{(z^2 + R^2)^{3/2} - z(\frac{3}{2})(z^2+R^2)^{1/2}(2z)}{(z^2 + R^2)^3}$. Setting the numerator to zero gives $(z^2 + R^2) - 3z^2 = 0$, which simplifies to $R^2 - 2z^2 = 0$. Solving for $z$ yields $z = R/\sqrt{2}$.
3. Consider two situations. Situation 1: A point charge $+Q$ creates an electric field of magnitude $E_1$ at a distance $R$ from the charge. Situation 2: A conducting sphere of radius $R_0$ with net charge $+Q$ creates an electric field of magnitude $E_2$ at a distance $R$ from its surface. How do $E_1$ and $E_2$ compare, assuming $R>0$ and $R_0>0$?
   Answer: $E_1 > E_2$ In Situation 1, the electric field is $E_1 = kQ/R^2$. In Situation 2, the electric field outside a conducting sphere is the same as that of a point charge located at its center. The distance from the center to the point in question is $R + R_0$. Thus, $E_2 = kQ/(R+R_0)^2$. Since $R+R_0 > R$, the denominator of $E_2$ is larger than the denominator of $E_1$, which means $E_2 < E_1$.
4. A point charge of $+2.0 \text{ nC}$ is located at the origin, and a point charge of $-2.0 \text{ nC}$ is located at $x = 6.0 \text{ m}$. What is the magnitude of the net electric field at $x = 3.0 \text{ m}$?
   Answer: $4.0 \text{ N/C}$ At the midpoint ($x = 3.0 \text{ m}$), the distance to each charge is $r=3.0 \text{ m}$. The field from the positive charge points in the +x direction. The field from the negative charge also points in the +x direction. The net field is the vector sum. $E_{net} = E_{pos} + E_{neg} = \frac{k|q_{pos}|}{r^2} + \frac{k|q_{neg}|}{r^2} = 2 \frac{(9.0 \times 10^9 \text{ N m}^2/\text{C}^2)(2.0 \times 10^{-9} \text{ C})}{(3.0 \text{ m})^2} = 4.0 \text{ N/C}$.
5. A thin spherical shell has radius $0.40\ \text{m}$ and total charge $+5.0\ \mu\text{C}$. Based on the scenario described, what is the electric field magnitude at $r=0.20\ \text{m}$ ($\varepsilon_0=8.85\times10^{-12}$)?
   Answer: $E=0\ \text{N/C}$ This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding electric fields for spherical shells and applying Gauss's Law. For a thin spherical shell with all charge on its surface at radius R, the electric field is zero everywhere inside (r < R) due to the shell theorem. In this scenario, we have a shell of radius 0.40 m and need the field at r = 0.20 m, which is inside the shell. Choice A is correct because it recognizes that E = 0 for any point inside a uniformly charged spherical shell, regardless of the charge magnitude or the specific location inside. Choice B is incorrect because it calculates the field as if the point were outside the shell or as if all charge were at the center. To help students: Emphasize the shell theorem - inside any spherical shell of charge, the net field from all charge elements cancels to zero. Watch for: students incorrectly applying the point charge formula inside shells or confusing this with solid spheres.
6. A thin rod of length $L$ lies along the x-axis from $x=a$ to $x=a+L$. The rod has a total charge $+Q$ distributed uniformly. Which integral correctly represents the magnitude of the electric field at the origin ($x=0$)?
   Answer: $E = \int_a^{a+L} \frac{k(Q/L)}{x^2} dx$ The linear charge density is $\lambda = Q/L$. A differential element of charge is $dq = \lambda dx = (Q/L)dx$. This element is at a distance $x$ from the origin. The differential field it produces at the origin is $dE = \frac{k dq}{x^2} = \frac{k(Q/L)dx}{x^2}$. To find the total field, this expression must be integrated over the entire length of the rod, which corresponds to the limits $x=a$ to $x=a+L$.
7. Which of the following statements about electric field lines is correct?
   Answer: The tangent to an electric field line at any point gives the direction of the electric field at that point. By convention, the direction of the electric field vector at any point in space is tangent to the electric field line passing through that point. Field lines cannot cross, as this would imply two different field directions at a single point. Field lines can originate from or terminate at infinity. The spacing indicates field strength, which varies for a point charge.
8. An isolated, hollow, conducting sphere is given a net positive charge $+Q$. Which statement correctly describes the electric field associated with this sphere after it has reached electrostatic equilibrium?
   Answer: The electric field is zero everywhere inside the sphere and is perpendicular to the surface just outside the sphere. For a conductor in electrostatic equilibrium, the net electric field inside the conducting material must be zero. Since the sphere is hollow, the field in the cavity is also zero by Gauss's law. The excess charge resides on the outer surface. Electric field lines must be perpendicular to the surface of a conductor, which is an equipotential surface.
9. A proton is released from rest in a uniform electric field of magnitude $1.0 \times 10^5 \text{ N/C}$. What is the magnitude of its initial acceleration?
   Answer: $9.6 \times 10^{12} \text{ m/s}^2$ The force on the proton is $F = qE$. Using Newton's second law, $F = ma$, the acceleration is $a = F/m = qE/m$. Using the charge of a proton ($1.60 \times 10^{-19} \text{ C}$) and the mass of a proton ($1.67 \times 10^{-27} \text{ kg}$), we get $a = \frac{(1.60 \times 10^{-19} \text{ C})(1.0 \times 10^5 \text{ N/C})}{1.67 \times 10^{-27} \text{ kg}} \approx 9.6 \times 10^{12} \text{ m/s}^2$.
10. An electron is in a uniform electric field of magnitude $300 \text{ N/C}$ directed in the positive x-direction. What is the force on the electron?
    Answer: $4.8 \times 10^{-17} \text{ N}$ in the negative x-direction. The electric force is given by the equation $\vec{F} = q\vec{E}$. For an electron, the charge $q$ is negative ($-e = -1.60 \times 10^{-19} \text{ C}$). Therefore, the force vector is in the opposite direction to the electric field vector. The magnitude of the force is $F = |q|E = (1.60 \times 10^{-19} \text{ C})(300 \text{ N/C}) = 4.8 \times 10^{-17} \text{ N}$. The direction is the negative x-direction.
11. An uncharged, isolated conducting sphere is placed in a pre-existing uniform external electric field that points to the right. After electrostatic equilibrium is reached, which statement best describes the electric field inside the sphere?
    Answer: The electric field is zero because induced charges create an internal field that cancels the external field. When a conductor is placed in an external electric field, its free charges redistribute. Negative charges move to the side opposing the field (left side), and positive charges are left on the other side (right side). This separation of induced charges creates an internal electric field that points to the left. The charges continue to move until this internal field perfectly cancels the external field everywhere inside the conductor, resulting in a net electric field of zero.
12. An electric dipole consists of two point charges of equal magnitude and opposite sign separated by a distance $d$. For points far from the dipole ($r \gg d$), how does the magnitude of the electric field $E$ on the perpendicular bisector of the dipole depend on the distance $r$ from the center of the dipole?
    Answer: $E \propto 1/r^3$ The electric fields from the two charges partially cancel. The net field is proportional to the dipole moment $p=qd$. A detailed derivation shows that for points far away on the perpendicular bisector, the electric field magnitude is given by $E \approx \frac{kp}{r^3}$. The fields from the individual charges, which are proportional to $1/r^2$, nearly cancel, and the remaining net field falls off more rapidly, as $1/r^3$.
13. A positive test charge is placed at rest at point P in a region described by electric field lines. Assume only the electric force acts on the charge. Which statement best describes the initial motion of the charge?
    Answer: It will accelerate in the direction tangent to the field line at P. The electric field vector at any point is defined as being tangent to the electric field line at that point. The force on a positive charge is in the same direction as the electric field, $\vec{F} = q\vec{E}$. Since there is a net force, the charge will accelerate (not move at constant velocity). The initial acceleration is therefore tangent to the field line. Its subsequent path may curve and not follow the field line exactly.
14. Two large, parallel, non-conducting plates are separated by a small distance. The plate on the left has a uniform positive surface charge density $+\sigma$, and the plate on the right has $-\sigma$. What are the magnitudes of the electric field between the plates ($E_{in}$) and to the right of the right plate ($E_{out}$)?
    Answer: $E_{in} = \sigma/\epsilon_0$, $E_{out} = 0$ Each infinite plate creates a uniform electric field of magnitude $\sigma/(2\epsilon_0)$. Between the plates, the field from the positive plate and the field from the negative plate both point in the same direction (to the right). By superposition, they add to $E_{in} = \sigma/(2\epsilon_0) + \sigma/(2\epsilon_0) = \sigma/\epsilon_0$. To the right of the right plate, the field from the positive plate points right, and the field from the negative plate points left. They are equal in magnitude and cancel out, so $E_{out} = 0$.
15. A point charge $+q$ is located at the origin. A second point charge $-2q$ is located on the x-axis at $x=d$. At which point on the x-axis, other than at infinity, is the net electric field equal to zero?
    Answer: $x = -(\sqrt{2}+1)d$ The fields can only cancel in a region where they point in opposite directions. This occurs for $x<0$. Let the point be at position $x$. The field from $+q$ points left, and the field from $-2q$ points right. For the magnitudes to be equal: $\frac{k|q|}{x^2} = \frac{k|-2q|}{(d-x)^2}$. Since $x<0$, this is $\frac{q}{x^2} = \frac{2q}{(d+|x|)^2}$. Letting $x' = |x|$, we have $(d+x')^2 = 2x'^2$. Taking the square root gives $d+x' = \sqrt{2}x'$ (we need the positive root for magnitudes). Solving for $x'$ gives $x' = d/(\sqrt{2}-1) = d(\sqrt{2}+1)$. Since $x = -x'$, the position is $x = -d(\sqrt{2}+1)$.
16. A student correctly states that the electric field inside a charged conductor in electrostatic equilibrium is zero. The student then claims that because the electric flux through any closed surface inside the conductor is zero, Gauss's law implies there can be no net charge enclosed by such a surface. Evaluate the student's reasoning.
    Answer: The reasoning is correct; a zero field implies zero flux, which in turn implies zero enclosed charge. The student's reasoning is entirely correct. In electrostatic equilibrium, the electric field within the material of a conductor is zero. If $\vec{E}=0$ everywhere on a closed surface, the electric flux integral $\oint \vec{E} \cdot d\vec{A}$ is zero. By Gauss's Law, $\oint \vec{E} \cdot d\vec{A} = Q_{enc}/\epsilon_0$. Therefore, a zero flux implies that the net charge enclosed, $Q_{enc}$, must be zero. This is why any net charge on a conductor resides on its surface.
17. At a specific point P in space, the net electric field is determined to be zero. Which of the following statements must be true?
    Answer: A test charge placed at point P will experience zero net electric force. The definition of the electric field is the electric force per unit charge, $\vec{E} = \vec{F}/q$. If the electric field $\vec{E}$ at point P is zero, the net electric force $\vec{F}$ on any test charge $q$ placed at that point must also be zero. The potential may be non-zero, charges can be arranged to create a zero field at a point, and the equilibrium could be unstable.
18. An infinite plane has uniform surface charge density $\sigma=-4.0\times10^{-6}\ \text{C/m}^2$. Based on the scenario described, what is the electric field magnitude and direction $0.30\ \text{m}$ above the plane ($\varepsilon_0=8.85\times10^{-12}$)?
    Answer: $E=2.3\times10^{5}\ \text{N/C}$, toward the plane This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding electric fields due to infinite charged planes using Gauss's Law. For an infinite plane with uniform surface charge density σ, the electric field has constant magnitude E = |σ|/(2ε₀) at any distance from the plane, with direction determined by the sign of σ. In this scenario, we have σ = -4.0×10⁻⁶ C/m², so the field magnitude is E = (4.0×10⁻⁶)/(2×8.85×10⁻¹²) = 2.26×10⁵ ≈ 2.3×10⁵ N/C. Choice A is correct because it has the right magnitude and, since σ is negative, the field points toward the plane (negative charges attract positive test charges). Choice C is incorrect because it has the wrong direction - negative surface charge creates fields pointing toward the plane, not away. To help students: Emphasize that for infinite planes, field magnitude is independent of distance and depends only on σ. Watch for: sign errors in determining field direction and forgetting the factor of 2 in the denominator.
19. Four identical positive point charges $+Q$ are fixed at the corners of a square. What is the direction of the net electric field at the exact center of the square?
    Answer: The net electric field is zero. Due to the symmetry of the arrangement, the electric field vector created by each charge at the center of the square is equal in magnitude and directed away from that charge. The field from the charge at one corner is exactly canceled by the field from the charge at the diagonally opposite corner. The vector sum of all four fields at the center is zero.
20. A point charge $Q=+3.0\ \mu\text{C}$ is fixed in vacuum. Based on the scenario described, calculate the electric field magnitude at $r=0.40\ \text{m}$ using Gauss's Law ($k=8.99\times10^{9}$).
    Answer: $E=1.7\times10^{5}\ \text{N/C}$, radially outward This question tests AP Physics C: Electricity and Magnetism skills, specifically calculating electric fields from point charges using Coulomb's law (which is equivalent to applying Gauss's Law to spherical symmetry). For a point charge Q in vacuum, the electric field at distance r is E = kQ/r², where k = 1/(4πε₀) = 8.99×10⁹ N·m²/C². In this scenario, Q = +3.0 μC and r = 0.40 m, so E = (8.99×10⁹)(3.0×10⁻⁶)/(0.40)² = 1.69×10⁵ ≈ 1.7×10⁵ N/C. Choice A is correct because it correctly calculates the magnitude and, since Q is positive, the field points radially outward from the charge. Choice B is incorrect because it has the wrong direction - positive charges create outward-pointing fields. To help students: Reinforce that field direction depends on charge sign: positive charges create outward fields, negative charges create inward fields. Watch for: calculation errors with scientific notation and confusion about field direction.

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