Let's solve each part step by step: --- ## (a) Direction of Electric Field in Diagram A **Between the plates in diagram A, draw a single vector indicating the direction of the electric field.** - The electric field **direction** is from higher potential (+100 V) to lower potential ( V). - Therefore, the electric field points **downward** (from the top plate to the bottom plate). --- ## (b) Student's Statement > "If the electrons are released from rest, they will move in the direction of the electric field." **Is this statement correct?** **Incorrect.** **Explanation:** - The electric field direction is the direction a **positive** charge would move. - Electrons are **negative** charges, so they move **opposite** to the direction of the electric field. **Corrected Statement:** *If electrons are released from rest, they will move in the direction **opposite** to the electric field.* --- ## (c) Rank the Magnitude of the Electric Field The electric field between parallel plates: \[ E = \frac{V}{d} \] where \( V \) is the voltage difference and \( d \) is the separation. Let's calculate for each case: 1. **A:** \( V = 100\,V, \quad d = 2\,cm = .02\,m \) \( E_A = \frac{100}{.02} = 500\,\text{V/m} \) 2. **B:** \( V = 100\,V, \quad d = 1\,cm = .01\,m \) \( E_B = \frac{100}{.01} = 10{,}000\,\text{V/m} \) 3. **C:** \( V = 50\,V, \quad d = 1\,cm = .01\,m \) \( E_C = \frac{50}{.01} = 500\,\text{V/m} \) 4. **D:** \( V = 100\,V, \quad d = 1\,cm = .01\,m \) \( E_D = \frac{100}{.01} = 10{,}000\,\text{V/m} \) **Ranking (from greatest to least):** - B and D (10,000 V/m) > A and C (5,000 V/m) - [B D] > [A C] --- ## (d) Rank Final Speeds of Electrons Electrons are released from rest. The final speed depends on the **potential difference the electron travels through**: \[ \frac{1}{2}mv^2 = q \Delta V \] - \( m \) = mass of electron - \( q = -e \) = charge of electron (magnitude is \( e \)) - \( \Delta V \) = potential difference traversed (from start to plate it ends up at) Let's check each case: ### **A:** - Electron starts halfway between (1 cm from each plate), so \(\Delta V = 50\,V\) (from 50 V to V or 50 V to 100 V). - When it reaches the bottom plate, it has traversed 50 V. ### **B:** - Electron starts halfway between (.5 cm from each plate), so \(\Delta V = 50\,V\). ### **C:** - Electron starts halfway between (.5 cm from each plate), so \(\Delta V = 25\,V\). ### **D:** - Electron starts .75 cm above the bottom plate (closer to the bottom plate). Plate separation is 1 cm, so it's .25 cm from the top plate. - Let's find \(\Delta V\) from starting point to bottom plate ( V): - The potential drops linearly from 100 V (top) to V (bottom) over 1 cm. - At .75 cm above the bottom plate, potential = \( \frac{.25}{1} \times 100\,V = 25\,V \) above V. - So, electron moves through \( 25\,V \) from starting point to bottom plate. **Ranking (from greatest to least):** - A and B (50 V) > C and D (25 V) - [A B] > [C D] --- ### **Final Answers:** --- ### (a) **Electric field direction:** Downward (from +100 V to V). --- ### (b) **Statement:** Incorrect. **Correction:** Electrons move opposite to the direction of the electric field. --- ### (c) **Electric field magnitude ranking:** [B D] > [A C] --- ### (d) **Final electron speed ranking:** [A B] > [C D] **Explanation:** The final speed depends on the potential difference the electron moves through, not the electric field strength alone. A and B have the largest potential difference (50 V), so their electrons will achieve the greatest speed; C and D have 25 V, so their electrons will have lower final speeds.