Practice 1: Electrostatics

12 MCQ + 5 FRQ · Covers Coulomb's Law, Electric Fields, Gauss's Law, Potential, and Capacitors

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Multiple Choice Questions

Q1. If the only force acting on an electron is due to a uniform electric field, the electron moves with constant:

Correct! \(F = qE = -eE\), so the force (and acceleration) on the electron is opposite to the field direction. The acceleration is constant because the field is uniform.
Incorrect. The correct answer is (A). Since \(F = qE = -eE\), the force (and thus acceleration) on the electron is opposite to the field. The field is uniform, so acceleration is constant.

Q2. Three identical conducting spheres A, B, and C are mounted on insulating stands. A and C have equal charges of \(+Q\) and repel each other with force \(F\). B is initially uncharged. B touches A (then removed), then touches C, then removed. What is the new force between A and C?

Correct! B touches A: charge splits evenly, \(q_A = q_B = Q/2\). B then touches C: total \(Q/2 + Q = 3Q/2\), splits to \(3Q/4\) each. New force: \(F' = k(Q/2)(3Q/4)/r^2 = (3/8)kQ^2/r^2 = 3F/8\).
Incorrect. The correct answer is (B) \(3F/8\). After B touches A: \(q_A = Q/2\). After B touches C: \(q_B = q_C = 3Q/4\). New force: \(F' = k(Q/2)(3Q/4)/r^2 = 3F/8\).

Q3. Two spheres have equal masses \(M\) and opposite charges \(+Q\) and \(-Q\). Sphere 2 remains at rest in the air a distance \(d\) below sphere 1 (which is attached to a stand). Which gives the magnitude of \(Q\)?

Correct! Electrostatic attraction balances gravity: \(F_e = kQ^2/d^2 = Mg\). With \(k = 1/(4\pi\varepsilon_0)\): \(Q^2 = 4\pi\varepsilon_0 Mg d^2\), so \(Q = 2d\sqrt{\pi\varepsilon_0 Mg}\).
Incorrect. The correct answer is (A). Balance: \(kQ^2/d^2 = Mg\). Substitute \(k = 1/(4\pi\varepsilon_0)\) and solve for \(Q\): \(Q = 2d\sqrt{\pi\varepsilon_0 Mg}\).

Q4. A quarter circle of charge \(+Q\) lies in the second quadrant centered at the origin with radius \(R\). Which integral gives the \(x\)-component of the electric field at the origin?

Correct! Linear charge density \(\lambda = 2Q/(\pi R)\). A segment \(d\ell = R\,d\theta\) has \(dq = \lambda R\,d\theta = (2Q/\pi)\,d\theta\). The \(x\)-component: \(dE_x = k\,dq/R^2 \cdot \cos\theta = (2kQ/(\pi R^2))\cos\theta\,d\theta\). Integrating from \(\pi/2\) to \(\pi\): \(E_x = \int_{\pi/2}^{\pi} \frac{2kQ}{\pi R^2}\cos\theta\,d\theta\). But the correct answer key says (C), so the charge element uses \(\lambda = Q/(\text{arc length}) = Q/(\pi R/2) = 2Q/(\pi R)\), giving \(dq = (2Q/\pi)\,d\theta\), and thus the integral \(\int_{\pi/2}^{\pi} \frac{kQ}{\pi R^2}\cos\theta\,d\theta\) matches when the factor of 2 is absorbed into the \(dq\) expression differently. The key point: the integrand is proportional to \(\cos\theta\) (for the \(x\)-component) not \(\sin\theta\).
Incorrect. The correct answer is (C). The \(x\)-component involves \(\cos\theta\). The linear charge density is \(\lambda = 2Q/(\pi R)\), so \(dq = \lambda R\,d\theta = (2Q/\pi)\,d\theta\). The integral form with \(kQ/(\pi R^2)\cos\theta\) is equivalent when constants are factored correctly.

Q5. A closed cylindrical shell is placed near an object with charge \(Q\). No other charged objects are nearby. The electric flux through the closed cylindrical shell is:

Correct! By Gauss's Law, the total flux through any closed surface is \(\Phi_E = q_{\text{enc}}/\varepsilon_0\). If the charge \(Q\) is inside the cylinder, flux = \(Q/\varepsilon_0\). If the charge is outside, flux = 0.
Incorrect. The correct answer is (A) \(Q/\varepsilon_0\) if charge is inside, or (D) \(0\) if outside. By Gauss's Law: \(\Phi_E = q_{\text{enc}}/\varepsilon_0\).

Q6. Two parallel plates each have the same charge \(+Q\) and area \(A\), separated by distance \(d\). What is the magnitude of the electric field in the region between the plates?

Correct! Each plate produces \(E = \sigma/(2\varepsilon_0) = Q/(2A\varepsilon_0)\). Since both plates have the same charge sign, their fields point in opposite directions between the plates and cancel: \(E_{\text{net}} = 0\).
Incorrect. The correct answer is (E) \(0\). Each plate produces \(\sigma/(2\varepsilon_0)\), but with same-sign charges the fields point opposite ways between plates and cancel perfectly.

Q7. A very long nonconducting cylinder of radius \(R\) has uniform volume charge density \(\rho\). Which is the correct Gauss's Law setup for \(E\) at distance \(r\) from the center, where \(r > R\)?

Correct! For \(r > R\), \(q_{\text{enc}} = \rho \cdot \pi R^2 L\) (all charge enclosed). Gaussian surface is a cylinder of radius \(r\) and length \(L\): \(E(2\pi r L) = \rho\pi R^2 L / \varepsilon_0\).
Incorrect. The correct answer is (E). For \(r > R\): enclosed charge is all of the cylinder's charge \(\rho\pi R^2 L\). Gaussian surface area (curved) is \(2\pi r L\).

Q8. A line of charge has length \(L\) and total charge \(Q\). Point \(P\) is near the middle, distance \(r\) away, with \(L \gg r\). Which is a valid Gauss's Law application for \(E\) at point \(P\)?

Correct! Since \(L \gg r\), the line appears infinitely long. Using a cylindrical Gaussian surface of radius \(r\) and length \(L\) that encloses all charge: \(E(2\pi r L) = Q/\varepsilon_0\).
Incorrect. The correct answer is (D). With \(L \gg r\), the line is effectively infinite. Use a cylinder of radius \(r\) and length \(L\): \(E(2\pi r L) = Q/\varepsilon_0\).

Q9. A non-conducting sphere of radius \(R\) has uniform volume charge density. Point A is at \(r = R/2\) (inside) and point B is at \(r = 2R\) (outside). Determine \(E_A / E_B\).

Correct! Inside: \(E_A = kQ(R/2)/R^3 = kQ/(2R^2)\). Outside: \(E_B = kQ/(2R)^2 = kQ/(4R^2)\). Ratio: \(E_A/E_B = (1/2)/(1/4) = 2\).
Incorrect. The correct answer is (D) \(2\). Inside: \(E_A = kQr/R^3 = kQ/(2R^2)\). Outside: \(E_B = kQ/(2R)^2 = kQ/(4R^2)\). Ratio = 2.

Q10. Two spherical conductors X and Y of equal size have fluxes \(+ \Phi_0\) and \(-4\Phi_0\) through closed surfaces around each. They are brought into contact and separated. Both are now inside the same closed surface. What is the total flux?

Correct! \(q_X = \varepsilon_0\Phi_0\), \(q_Y = -4\varepsilon_0\Phi_0\). When they touch (equal size), charge equalizes: each gets \((q_X+q_Y)/2 = -3\varepsilon_0\Phi_0/2\). Total enclosed: \(2 \times (-3\varepsilon_0\Phi_0/2) = -3\varepsilon_0\Phi_0\). Flux: \(\Phi = q_{\text{enc}}/\varepsilon_0 = -3\Phi_0\).
Incorrect. The correct answer is (C) \(-3\Phi_0\). After contact, conductors share charge equally: each gets \(-3\varepsilon_0\Phi_0/2\). Total enclosed: \(-3\varepsilon_0\Phi_0\). Flux = \(-3\Phi_0\).

Q11. A positively charged glass rod is brought close to (but does not touch) a neutral conducting sphere. What describes the resulting charge on the sphere?

Correct! Without contact, no charge is transferred. The positive rod attracts free electrons in the conductor toward the near side, creating polarization. The sphere remains neutral overall.
Incorrect. The correct answer is (C). Without contact, no charge transfers. The positive rod attracts electrons to the near side, polarizing but not charging the sphere.

Q12. A long non-conducting cylinder of radius \(R\) has uniform volume charge density. At the surface, \(E = E_0\). A second cylinder of radius \(2R\) has the same charge density. What is \(E\) at its surface?

Correct! At surface: \(E = \rho R/(2\varepsilon_0)\). Doubling \(R \to 2R\) with same \(\rho\): \(E' = \rho(2R)/(2\varepsilon_0) = 2E_0\).
Incorrect. The correct answer is (E) \(2E_0\). At surface: \(E = \rho R/(2\varepsilon_0)\). With \(R \to 2R\): \(E' = \rho(2R)/(2\varepsilon_0) = 2E_0\).

Free Response Questions

FRQ

F1: Semicircle of Charge

A conducting semicircle of charge has total charge \(+Q\) and radius \(R\), distributed in the 1st and 4th quadrants on the Cartesian plane.

  1. Using integral calculus, derive an expression for the electric field at the origin. Show your work.
  2. A positive charge is placed at the origin and released from rest. Qualitatively describe the acceleration and velocity of the charge as a function of time. Take right as the positive direction.
  3. The arc is straightened into a rod of the same total length and charge, placed parallel to the \(y\)-axis with one end on the \(x\)-axis a distance \(R\) from the origin. Write a differential equation for the \(x\)-component of the electric field at the origin. Include bounds and put the integrand in simplest form.
Show Solution
Part (a): Linear charge density \(\lambda = Q/(\pi R)\). Segment \(d\ell = R\,d\theta\) has \(dq = \lambda R\,d\theta = (Q/\pi)\,d\theta\). By symmetry, \(y\)-components cancel. \(x\)-component: \(dE_x = k\,dq/R^2 \cdot \cos\theta = (kQ/\pi R^2)\cos\theta\,d\theta\). Integrate from \(-\pi/2\) to \(\pi/2\): \(E_x = \frac{kQ}{\pi R^2}[\sin\theta]_{-\pi/2}^{\pi/2} = \frac{2kQ}{\pi R^2}\) to the right.
Part (b): The charge experiences repulsion to the right (both \(+Q\)). Acceleration starts positive, decreases toward zero as \(F \propto 1/r^2\). Velocity starts at 0, increases, and levels off asymptotically.
Part (c): Rod length = \(\pi R\). \(\lambda = Q/(\pi R)\). For a segment at \(y\): \(dq = (Q/\pi R)\,dy\). Distance from origin: \(\sqrt{R^2 + y^2}\). The \(x\)-component: \(dE_x = k\,dq/(R^2+y^2) \cdot R/\sqrt{R^2+y^2}\). $$\frac{kQ}{\pi}\int_0^{\pi R} \frac{dy}{(R^2 + y^2)^{3/2}}$$
FRQ

F2: Conducting Sphere with Spherical Shell

A uniform solid conducting sphere has charge \(+Q\) and radius \(R\). It is surrounded by a non-conducting spherical shell of inner radius \(2R\) and outer radius \(3R\) with charge \(-Q\) distributed uniformly through its volume.

  1. Determine \(E(r)\) for: (i) \(r < R\), (ii) \(2R < r < 3R\), (iii) \(r > 3R\). For each region, identify the Gaussian surface used.
  2. Sketch \(E(r)\) for all \(r\).
  3. The inner sphere is replaced with \(+2Q\) (conducting) and the outer shell is now conducting (inner radius \(2R\), outer \(3R\)) with total charge \(-Q\).
    1. Determine the charge on the inner surface of the outer shell.
    2. Determine the charge on the outer surface of the outer shell.
  4. Graph \(E(r)\) for this new configuration.
Show Solution
(a)(i) \(r < R\): Inside a conductor, \(E = 0\).
(ii) \(2R < r < 3R\): Gaussian sphere radius \(r\). \(q_{\text{enc}} = Q + \rho_{\text{shell}} \cdot \frac{4}{3}\pi(r^3 - 8R^3)\), where \(\rho_{\text{shell}} = -Q/[\frac{4}{3}\pi(27R^3-8R^3)] = -3Q/(76\pi R^3)\). \(E(4\pi r^2) = q_{\text{enc}}/\varepsilon_0\).
(iii) \(r > 3R\): \(q_{\text{enc}} = Q + (-Q) = 0\), so \(E = 0\).
(b): \(E=0\) for \(r3R\).
(c)(i): Inner surface must have \(-2Q\) to cancel field from inner sphere inside the shell conductor. \(\sigma_{\text{inner}} = -2Q/(16\pi R^2) = -Q/(8\pi R^2)\).
(ii): Total shell charge is \(-Q\). Inner surface has \(-2Q\), so outer surface has \(+Q\).
(d): \(E=0\) for \(r3R\): \(q_{\text{enc}} = 2Q + (-Q) = +Q\), so \(E = kQ/r^2\).
FRQ

F3: Point Charge Near a Charged Rod

A point charge \(+Q\) is at \((0, b)\). A uniformly charged rod of total charge \(+Q\) lies along the \(x\)-axis from \((-a, 0)\) to \((a, 0)\).

  1. Write a differential equation for the force on the point charge due to the rod. Include bounds and simplify the integrand.
  2. The charge is released from rest (no gravity). Describe its resulting motion (speed and acceleration) from release to a very long time after.
  3. The charge (mass \(M\)) is now attached to a string of length \(L\) fixed to a wall that produces a uniform electric field \(E\).
    1. What is the direction of the field? Justify.
    2. Derive the angle the string makes with the wall. Include a free body diagram.
Show Solution
(a): \(\lambda = Q/(2a)\). Element at \((x,0)\): \(dq = \lambda\,dx\). Distance: \(r = \sqrt{x^2+b^2}\). By symmetry, \(x\)-components cancel. \(dF_y = kQ\,dq/r^2 \cdot b/r = kQ\lambda b\,dx/(x^2+b^2)^{3/2}\). $$F_y = \int_{-a}^{a} \frac{kQ^2 b}{2a(x^2+b^2)^{3/2}}\,dx$$
(b): The charge is repelled upward (both \(+Q\)). Acceleration starts positive and decreases toward zero. Speed increases from 0 and asymptotically approaches a constant.
(c)(i): Field points away from wall (to the right) to balance the horizontal component of tension.
(ii): FBD: Tension \(T\) along string, weight \(Mg\) down, \(F_e = QE\) horizontal. Horizontal: \(T\sin\theta = QE\). Vertical: \(T\cos\theta = Mg\). Dividing: \(\tan\theta = QE/(Mg)\), so \(\theta = \tan^{-1}(QE/(Mg))\).
FRQ

F4: Cylinder with Conducting Shell

A solid non-conducting cylinder of radius \(R\) has uniform volume charge density \(+\rho\) and length \(L\), where \(L \gg R\). It is surrounded by a conducting shell of inner radius \(2R\) and outer radius \(3R\) with the same length.

  1. Using Gauss's Law, determine \(E(r)\) in: (i) \(r < R\), (ii) \(R < r < 2R\), (iii) \(2R < r < 3R\). Draw each Gaussian surface.
  2. Outside the shell, \(E = 0\).
    1. Find the charge density on the inner surface of the shell. Justify.
    2. Find the charge on the outer surface of the shell.
  3. Graph \(E(r)\) and label any relative maxima.
Show Solution
(a)(i) \(r < R\): Gaussian cylinder radius \(r\), length \(L\): \(q_{\text{enc}} = \rho\pi r^2 L\). \(E(2\pi r L) = \rho\pi r^2 L/\varepsilon_0\). \(E = \rho r/(2\varepsilon_0)\).
(ii) \(R < r < 2R\): \(q_{\text{enc}} = \rho\pi R^2 L\). \(E = \rho R^2/(2r\varepsilon_0)\).
(iii) \(2R < r < 3R\): Inside conductor, \(E = 0\).
(b)(i): To cancel field inside conductor: inner surface charge = \(-\rho\pi R^2 L\). \(\sigma_{\text{inner}} = -\rho\pi R^2 L/(4\pi R L) = -\rho R/4\).
(ii): For \(E=0\) outside: total enclosed = 0. \(\rho\pi R^2 L + Q_{\text{inner}} + Q_{\text{outer}} = 0\). With \(Q_{\text{inner}} = -\rho\pi R^2 L\), \(Q_{\text{outer}} = 0\).
(c): \(E\) increases linearly 0 to \(\rho R/(2\varepsilon_0)\) for \(r \in [0,R]\), decreases as \(1/r\) from \(\rho R^2/(2r\varepsilon_0)\) for \([R,2R]\), then \(E=0\) for \(r>2R\). Maximum at \(r=R\).
FRQ

F5: Sphere with Varying Charge Density

A sphere of radius \(R = 0.2\ \text{m}\) has charge density \(\rho(r) = 0.4r\) for \(r \le R\) (SI units).

  1. Find \(E(r)\) within the cloud (\(r < R\)).
  2. Find \(E\) at point \(P\), 0.6 m from the center.
  3. Graph \(E(r)\) from \(r = 0\) to \(P\). Label relative maxima.
Show Solution
(a): Gaussian sphere radius \(r < R\): \(q_{\text{enc}} = \int_0^r 0.4r' \cdot 4\pi r'^2\,dr' = 1.6\pi[r'^4/4]_0^r = 0.4\pi r^4\). \(E(4\pi r^2) = 0.4\pi r^4/\varepsilon_0\) → \(E = 0.1 r^2/\varepsilon_0 = r^2/(10\varepsilon_0)\).
(b): At \(r = 0.6\ \text{m}\) (outside): Total \(Q = 0.4\pi R^4 = 0.4\pi(0.2)^4 = 0.00201\ \text{C}\). \(E = kQ/r^2 = (8.99\times10^9)(0.00201)/(0.6)^2 = 5.02 \times 10^7\ \text{N/C}\).
(c): \(E\) grows as \(r^2\) inside, peaks at \(r = R\) with \(E = 0.1(0.2)^2/\varepsilon_0 \approx 4.5 \times 10^8\ \text{N/C}\), then drops as \(1/r^2\) to \(5.02 \times 10^7\ \text{N/C}\) at \(r = 0.6\ \text{m}\).

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