FAD1022 CQ — Electrostatics Quiz Set

Course quiz problems covering work done in electric fields, charge motion in parallel plates, electric potential from point charges, charged particle acceleration, potential difference calculations, and potential energy changes in uniform fields.

[!info] Quiz Set Contents This collection contains 8 problems from course quizzes covering electrostatics fundamentals:

  • Work and potential energy ($W = qV$, $\Delta U = q\Delta V$)
  • Motion in uniform electric fields (travel time, acceleration)
  • Electric potential from point charges (scalar superposition)
  • Potential in uniform fields and direction relationships

[!tip] Quick Answer Key (4 Problems)

Q Answer Trap
1 $2.4 \times 10^{3}$ J None — straightforward calculation
2 $5.84 \times 10^{-9}$ s Using vertical separation (45 mm) instead of horizontal distance (90 mm)
3 $+1.8 \times 10^{6}$ V Using full separation instead of half-distance (0.05 m); calculating E-field (vector) instead of V (scalar)
4 $6.33 \times 10^{14}$ m/s² Using proton mass instead of electron mass

Question 1: Work Done Transferring Charge

Problem Statement: A charge of 20 C is transferred across two points with a potential difference of 120 V. Determine the work done.

Given:

  • Charge: $q = 20$ C
  • Potential difference: $V = 120$ V

Solution:

Step 1: Use the work formula for electric potential (L4 §2) $$W = qV$$

Step 2: Substitute values $$W = (20)(120) = 2400\ \text{J}$$

Step 3: Express in scientific notation $$W = 2.4 \times 10^{3}\ \text{J}$$

Answer: $2.4 \times 10^{3}$ J

[!tip] Trap None — this is a straightforward calculation testing the basic relationship $W = qV$.

Concept: Electric Potential — work done moving charge through potential difference.

Question 2: Proton Travel Time in Parallel Plates

Problem Statement: A proton travelling at $1.54 \times 10^{7}$ m/s enters between parallel plates (90 mm long, 45 mm separation) with $E = 2.69 \times 10^{4}$ N/C. Calculate the time to travel through the plates.

Given:

  • Proton velocity: $v = 1.54 \times 10^{7}$ m/s
  • Plate length (horizontal): $d = 90$ mm = $90 \times 10^{-3}$ m = $0.090$ m
  • Plate separation (vertical): $h = 45$ mm = $0.045$ m
  • Electric field: $E = 2.69 \times 10^{4}$ N/C

Solution:

Step 1: Identify the relevant distance The proton travels horizontally through the plates. The time depends only on the horizontal motion.

Step 2: Use constant velocity motion Since the electric field is vertical (perpendicular to the proton's initial horizontal velocity), the horizontal component of velocity remains constant (L3 §Case 3).

$$t = \frac{d}{v}$$

Step 3: Calculate $$t = \frac{90 \times 10^{-3}}{1.54 \times 10^{7}}$$

$$t = \frac{0.090}{1.54 \times 10^{7}}$$

$$t = 5.84 \times 10^{-9}\ \text{s}$$

Answer: $5.84 \times 10^{-9}$ s

[!danger] Critical Trap The electric field affects the vertical motion (causing acceleration), but NOT the horizontal travel time.

Wrong approach: Using the 45 mm separation would give $t = (45 \times 10^{-3})/(1.54 \times 10^{7}) = 2.92 \times 10^{-9}$ s — this is incorrect because the proton travels the length of the plates (90 mm), not the separation between them.

The vertical separation (45 mm) would only be relevant for calculating vertical deflection, not travel time.

Concepts:

  • L3 §Case 3 — perpendicular entry into uniform E-field
  • Horizontal motion: constant velocity $v_x = v_0$
  • Vertical motion: accelerated by $a_y = qE/m$

Question 3: Electric Potential at Midpoint

Problem Statement: Two point charges $+3.4$ μC and $+6.6$ μC are separated by 0.10 m. Find the electric potential at the midpoint.

Given:

  • Charge 1: $q_1 = +3.4$ μC = $+3.4 \times 10^{-6}$ C
  • Charge 2: $q_2 = +6.6$ μC = $+6.6 \times 10^{-6}$ C
  • Separation: $d = 0.10$ m
  • Midpoint distance from each charge: $r = d/2 = 0.05$ m

Solution:

Step 1: Recall electric potential formula (L4 §3) $$V = k\frac{Q}{r}$$

where $k = 8.99 \times 10^{9}$ N·m²/C² (Coulomb constant)

Step 2: Calculate total potential at midpoint Electric potential is a scalar quantity, so we simply add the contributions from both charges:

$$V_{\text{total}} = V_1 + V_2 = k\frac{q_1}{r} + k\frac{q_2}{r} = \frac{k}{r}(q_1 + q_2)$$

Step 3: Substitute values $$V_{\text{total}} = \frac{8.99 \times 10^{9}}{0.05}(3.4 \times 10^{-6} + 6.6 \times 10^{-6})$$

$$V_{\text{total}} = \frac{8.99 \times 10^{9}}{0.05}(10 \times 10^{-6})$$

$$V_{\text{total}} = (1.798 \times 10^{11})(10^{-5})$$

$$V_{\text{total}} = 1.798 \times 10^{6}\ \text{V}$$

$$V_{\text{total}} \approx +1.8 \times 10^{6}\ \text{V}$$

Answer: $+1.8 \times 10^{6}$ V

[!danger] Common Traps Trap 1: Using the full separation (0.10 m) instead of half-distance (0.05 m). The midpoint is 0.05 m from each charge.

Trap 2: Calculating electric field (vector) instead of potential (scalar). Electric field at midpoint would require vector addition (fields oppose and partially cancel since both charges are positive). Potential simply adds algebraically.

Trap 3: Forgetting that potential is a scalar. Some students try to subtract or use vector addition — this is wrong for electric potential.

Concepts:

  • Electric Potential — scalar quantity from point charges
  • Coulomb's Law — $k = 8.99 \times 10^{9}$ N·m²/C²
  • Scalar vs vector: V adds algebraically, $\vec{E}$ adds vectorially

Question 4: Electron Acceleration in E-field

Problem Statement: An electron at $1.04 \times 10^{7}$ m/s enters between plates with $E = 3.60 \times 10^{3}$ N/C. Find the acceleration.

Given:

  • Electron initial velocity: $v_0 = 1.04 \times 10^{7}$ m/s
  • Electric field: $E = 3.60 \times 10^{3}$ N/C
  • Electron charge: $e = 1.602 \times 10^{-19}$ C
  • Electron mass: $m_e = 9.11 \times 10^{-31}$ kg

Solution:

Step 1: Force on charge in electric field (L2 §2.3) $$F = qE$$

Step 2: Apply Newton's Second Law $$F = ma \Rightarrow a = \frac{F}{m} = \frac{qE}{m}$$

Step 3: Substitute electron values $$a = \frac{(1.602 \times 10^{-19})(3.60 \times 10^{3})}{9.11 \times 10^{-31}}$$

Step 4: Calculate $$a = \frac{5.767 \times 10^{-16}}{9.11 \times 10^{-31}}$$

$$a = 6.33 \times 10^{14}\ \text{m/s}^2$$

Answer: $6.33 \times 10^{14}$ m/s²

[!warning] Critical Trap Using proton mass instead of electron mass.

If you mistakenly use $m_p = 1.67 \times 10^{-27}$ kg (proton mass): $$a = \frac{5.767 \times 10^{-16}}{1.67 \times 10^{-27}} = 3.45 \times 10^{11}\ \text{m/s}^2$$

This gives a completely wrong answer — the proton is ~1836 times more massive than the electron, so its acceleration would be much smaller.

Remember:

  • Electron mass: $m_e = 9.11 \times 10^{-31}$ kg
  • Proton mass: $m_p = 1.67 \times 10^{-27}$ kg

[!note] Direction Note For an electron (negative charge), the acceleration is opposite to the electric field direction. The magnitude is what the question asks for.

Concepts:

  • L3 — charge acceleration in uniform E-field
  • Electron properties — charge and mass
  • Force on charge: $F = qE$
  • Acceleration: $a = qE/m$

[!tip] Quick Answer Key — Additional Questions (4 Problems)

Q Answer Trap
5 $5.84 \times 10^{-9}$ s Using vertical separation instead of horizontal distance
6 $6.33 \times 10^{14}$ m/s² Using incorrect mass or including initial velocity unnecessarily
7 $0.1$ V None — straightforward calculation
8 $1.9 \times 10^{-16}$ J Sign error on ΔV; moving with vs against field direction

Question 5: Proton Travel Time in Parallel Plates (Alternate)

Problem Statement: A proton travelling at a speed of $1.54 \times 10^{7}$ m/s enter the space between two parallel plates (90 mm long, 45 mm separation). The electric field strength between the plates is $2.69 \times 10^{4}$ N/C. Calculate the time taken for proton to travel perpendicularly through the space between the plates.

Given:

  • Proton velocity: $v = 1.54 \times 10^{7}$ m/s
  • Plate length (horizontal): $d = 90$ mm = $0.090$ m
  • Plate separation (vertical): $h = 45$ mm = $0.045$ m
  • Electric field: $E = 2.69 \times 10^{4}$ N/C
  • Charge of proton: $q_p = 1.602 \times 10^{-19}$ C
  • Mass of proton: $m_p = 1.67 \times 10^{-27}$ kg

Solution:

Step 1: Identify the motion type The proton enters perpendicular to the electric field (horizontal entry, vertical E-field). The horizontal motion is independent of the electric field (L3 §Case 3).

Step 2: Use constant velocity for horizontal motion $$t = \frac{d}{v}$$

Step 3: Calculate $$t = \frac{90 \times 10^{-3}}{1.54 \times 10^{7}} = \frac{0.090}{1.54 \times 10^{7}}$$

$$t = 5.84 \times 10^{-9}\ \text{s}$$

Answer: $5.84 \times 10^{-9}$ s

[!danger] Critical Trap The electric field does NOT affect horizontal travel time. Using the 45 mm separation would give the wrong answer.

The E-field affects vertical deflection only, which can be calculated separately using $a = qE/m$ if needed.

Concepts:

  • L3 — perpendicular entry into uniform E-field
  • Horizontal: constant velocity $v_x = v_0$
  • Vertical: accelerated motion $a_y = qE/m$

Question 6: Electron Acceleration in Electric Field (Alternate)

Problem Statement: An electron travelling at a speed of $1.04 \times 10^{7}$ m/s enter the space perpendicularly between two parallel metal planes. The electric field strength between the metal planes is $3.60 \times 10^{3}$ N/C. Calculate the acceleration of electron in the electric field.

Given:

  • Electron initial velocity: $v_0 = 1.04 \times 10^{7}$ m/s
  • Electric field: $E = 3.60 \times 10^{3}$ N/C
  • Charge of electron: $e = 1.602 \times 10^{-19}$ C
  • Mass of electron: $m_e = 9.11 \times 10^{-31}$ kg

Solution:

Step 1: Force on charge in electric field $$F = qE = (1.602 \times 10^{-19})(3.60 \times 10^{3})$$

$$F = 5.767 \times 10^{-16}\ \text{N}$$

Step 2: Apply Newton's Second Law $$a = \frac{F}{m} = \frac{qE}{m}$$

Step 3: Substitute electron mass $$a = \frac{5.767 \times 10^{-16}}{9.11 \times 10^{-31}}$$

$$a = 6.33 \times 10^{14}\ \text{m/s}^2$$

Answer: $6.33 \times 10^{14}$ m/s²

[!warning] Traps

  • Initial velocity is irrelevant — acceleration depends only on F and m, not on how fast the particle is moving
  • Must use electron mass ($9.11 \times 10^{-31}$ kg), not proton mass
  • Electron accelerates opposite to E-field direction (negative charge)

Concepts:

  • Electron properties — charge and mass
  • Force on charge: $F = qE$
  • Acceleration: $a = qE/m$

Question 7: Potential Difference from Work

Problem Statement: An amount of work of 2 J is required to move a charge of 20 C from point A to point B. Calculate the potential difference between point A and point B.

Given:

  • Work done: $W = 2$ J
  • Charge: $q = 20$ C

Solution:

Step 1: Recall the work-potential relationship (L4 §2) $$W = q\Delta V$$

Step 2: Solve for potential difference $$\Delta V = \frac{W}{q} = \frac{2}{20}$$

$$\Delta V = 0.1\ \text{V}$$

Answer: $0.1$ V

[!tip] Note This is the inverse of Question 1. In Q1 we calculated $W$ given $q$ and $V$; here we calculate $\Delta V$ given $W$ and $q$.

Concept: Electric Potential — relationship between work, charge, and potential difference.

Question 8: Change in Potential Energy in Uniform Field

Problem Statement: A uniform electric field, with a magnitude of 600 N/C, is directed parallel to the positive x-axis. If the potential at $x = 3.0$ m is 1000 V, what is the change in potential energy of a proton as it moves from $x = 3.0$ m to $x = 1.0$ m?

Given:

  • Electric field: $E = 600$ N/C (along +x direction)
  • Potential at $x = 3.0$ m: $V_{3.0} = 1000$ V
  • Initial position: $x_i = 3.0$ m
  • Final position: $x_f = 1.0$ m
  • Proton charge: $q_p = 1.6 \times 10^{-19}$ C

Solution:

Step 1: Determine the potential at $x = 1.0$ m The electric field points in the direction of decreasing potential. Moving from $x = 3.0$ to $x = 1.0$ means moving opposite to E (negative x direction), so potential increases.

$$\Delta x = 3.0 - 1.0 = 2.0\ \text{m}$$

$$V_{1.0} = V_{3.0} + E \cdot \Delta x = 1000 + (600)(2.0)$$

$$V_{1.0} = 1000 + 1200 = 2200\ \text{V}$$

Step 2: Calculate potential difference $$\Delta V = V_{1.0} - V_{3.0} = 2200 - 1000 = 1200\ \text{V}$$

Step 3: Calculate change in potential energy $$\Delta U = q \cdot \Delta V = (1.6 \times 10^{-19})(1200)$$

$$\Delta U = 1.92 \times 10^{-16}\ \text{J}$$

Answer: $1.9 \times 10^{-16}$ J

[!danger] Critical Trap — Direction Matters Key relationship: Electric field points from high potential to low potential.

  • Moving with E (in direction of field): potential decreases
  • Moving against E (opposite to field): potential increases

Here, the proton moves from $x = 3.0$ to $x = 1.0$, which is opposite to the +x direction of E, so it moves to higher potential.

Wrong approach: Using $\Delta V = -E \cdot d$ gives $-1200$ V, leading to negative $\Delta U$. While the magnitude is correct, the sign indicates the proton gains potential energy (external work is done to move it against the field).

Concepts:

Summary Table — All Questions

Q Problem Type Key Formula Main Trap Answer
1 Work from potential difference $W = qV$ None — straightforward $2.4 \times 10^{3}$ J
2 Travel time in parallel plates $t = d/v$ Using vertical instead of horizontal distance $5.84 \times 10^{-9}$ s
3 Electric potential at midpoint $V = kQ/r$ (scalar sum) Using full distance; confusing V with E-field $+1.8 \times 10^{6}$ V
4 Electron acceleration $a = qE/m$ Using proton mass $6.33 \times 10^{14}$ m/s²
5 Proton travel time (alternate) $t = d/v$ Using vertical separation $5.84 \times 10^{-9}$ s
6 Electron acceleration (alternate) $a = qE/m$ Including velocity unnecessarily $6.33 \times 10^{14}$ m/s²
7 Potential difference from work $\Delta V = W/q$ None — straightforward $0.1$ V
8 PE change in uniform field $\Delta U = q\Delta V = qEd$ Direction of field vs motion $1.9 \times 10^{-16}$ J

Key Formulas

Work and Energy: $$W = qV = q\Delta V$$

Electric Potential: $$V = k\frac{Q}{r}$$ $$V_{\text{total}} = \sum_i V_i = k\sum_i \frac{Q_i}{r_i} \quad\text{(scalar sum)}$$

Motion in Uniform Electric Field: $$F = qE$$ $$a = \frac{qE}{m}$$ $$t = \frac{d}{v_x} \quad\text{(horizontal travel time)}$$

Constants:

Constant Symbol Value
Coulomb constant $k$ $8.99 \times 10^{9}$ N·m²/C²
Elementary charge $e$ $1.602 \times 10^{-19}$ C
Electron mass $m_e$ $9.11 \times 10^{-31}$ kg
Proton mass $m_p$ $1.67 \times 10^{-27}$ kg

Related