Question

The electric field \( \vec{E} \) in the region between the plates is:

• A. \( \left(2 \times 10^2 \, \frac{V}{m}\right) \hat{k} \)
• B. \( -\left(2 \times 10^2 \, \frac{V}{m}\right) \hat{k} \)
• C. \( \left(2 \times 10^4 \, \frac{V}{m}\right) \hat{k} \)
• D. \( -\left(2 \times 10^4 \, \frac{V}{m}\right) \hat{k} \)
• A. \( -\left(3.5 \times 10^{15} \, \text{m s}^{-2}\right) \hat{k} \)
• B. \( \left(3.5 \times 10^{15} \, \text{m s}^{-2}\right) \hat{k} \)
• C. \( \left(3.5 \times 10^{13} \, \text{m s}^{-2}\right) \hat{i} \)
• D. \( -\left(3.5 \times 10^{13} \, \text{m s}^{-2}\right) \hat{i} \)
• A. \( 9.0 \times 10^{-9} \, \text{s} \)
• B. \( 1.67 \times 10^{-8} \, \text{s} \)
• C. \( 1.67 \times 10^{-9} \, \text{s} \)
• D. \( 2.17 \times 10^{-9} \, \text{s} \)
• A. 10 mm
• B. 4.9 mm
• C. 5.9 mm
• D. 3.0 mm
• A. a
• B. b
• C. c
• D. d

Answer 1

The Correct Option is C

Concept: For parallel plates: \[ E = \frac{V}{d} \] Direction: Electric field points from higher potential plate to lower potential plate.
Step 1: Calculate magnitude. \[ V = 200 \, \text{V}, \quad d = 0.01 \, \text{m} \] \[ E = \frac{200}{0.01} = 2 \times 10^4 \, \text{V/m} \]
Step 2: Determine direction. From the figure, field is along +z direction. Unit vector along z-axis is \( \hat{k} \). Final Answer: \[ \vec{E} = 2 \times 10^4 \, \hat{k} \, \text{V/m} \]

Answer 2

Concept: Force on a charge in an electric field: \[ \vec{F} = q\vec{E} \] Acceleration: \[ \vec{a} = \frac{q\vec{E}}{m} \] For electron: \[ q = -e \]
Step 1: Electric field. From previous result: \[ \vec{E} = 2 \times 10^4 \, \hat{k} \, \text{V/m} \]
Step 2: Use electron charge and mass. \[ e = 1.6 \times 10^{-19} \, \text{C}, \quad m_e = 9.1 \times 10^{-31} \, \text{kg} \] \[ a = \frac{eE}{m} = \frac{1.6 \times 10^{-19} \times 2 \times 10^4}{9.1 \times 10^{-31}} \]
Step 3: Calculate magnitude. \[ a = \frac{3.2 \times 10^{-15}}{9.1 \times 10^{-31}} \approx 3.5 \times 10^{15} \, \text{m s}^{-2} \]
Step 4: Direction. Electron has negative charge, so acceleration is opposite to field. Field is along \( +\hat{k} \) → acceleration along \( -\hat{k} \). Final Answer: \[ \vec{a} = -3.5 \times 10^{15} \, \hat{k} \, \text{m s}^{-2} \]

Answer 3

Concept: The electron enters horizontally between the plates. Electric field acts vertically, so horizontal motion remains uniform. Time inside plates depends only on horizontal velocity.
Step 1: Given data. \[ \text{Plate length} = 0.05 \, \text{m} \] \[ v_x = 3 \times 10^7 \, \text{m/s} \]
Step 2: Time of travel. \[ t = \frac{\text{distance}}{\text{velocity}} = \frac{0.05}{3 \times 10^7} \]
Step 3: Calculate. \[ t = \frac{5 \times 10^{-2}}{3 \times 10^7} = 1.67 \times 10^{-9} \, \text{s} \] Final Answer: \[ t = 1.67 \times 10^{-9} \, \text{s} \]

Answer 4

Concept: Electron experiences vertical acceleration due to electric field, while horizontal motion is uniform. Vertical displacement: \[ y = \frac{1}{2} a t^2 \]
Step 1: Known values. \[ a = 3.5 \times 10^{15} \, \text{m s}^{-2} \] \[ t = 1.67 \times 10^{-9} \, \text{s} \]
Step 2: Substitute into formula. \[ y = \frac{1}{2} \times 3.5 \times 10^{15} \times (1.67 \times 10^{-9})^2 \]
Step 3: Calculate. \[ (1.67 \times 10^{-9})^2 = 2.79 \times 10^{-18} \] \[ y = 0.5 \times 3.5 \times 2.79 \times 10^{-3} \] \[ y \approx 4.9 \times 10^{-3} \, \text{m} \]
Step 4: Convert to mm. \[ y = 4.9 \, \text{mm} \] Final Answer: \[ \boxed{4.9 \, \text{mm}} \]

Answer 5

Concept: Electron enters horizontally with velocity along x-axis and experiences:

No force in horizontal direction → uniform motion
Constant vertical acceleration due to electric field
This produces projectile-like motion.
Step 1: Nature of motion. The motion is similar to:

Uniform velocity in x-direction
Uniform acceleration in vertical direction
Hence, trajectory is a parabola.
Step 2: Direction of deflection. From earlier results:

Electric field is along \( +\hat{k} \)
Electron (negative charge) accelerates opposite → downward

Step 3: Identify correct path. The path should:

Start horizontally
Curve downward gradually (parabolic path)
Among the options, only path c shows downward curvature. Final Answer: Path c