Question

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

• A. \(\left(2\times10^{2}\,\dfrac{\mathrm{V}}{\mathrm{m}}\right)\hat{k}\)
• B. \(-\left(2\times10^{2}\,\dfrac{\mathrm{V}}{\mathrm{m}}\right)\hat{k}\)
• C. \(\left(2\times10^{4}\,\dfrac{\mathrm{V}}{\mathrm{m}}\right)\hat{k}\)
• D. \(-\left(2\times10^{4}\,\dfrac{\mathrm{V}}{\mathrm{m}}\right)\hat{k}\)
• A. \(-\left(3.5\times10^{15}\,\mathrm{m\,s^{-2}}\right)\hat{k}\)
• B. \(\left(3.5\times10^{15}\,\mathrm{m\,s^{-2}}\right)\hat{k}\)
• C. \(\left(3.5\times10^{13}\,\mathrm{m\,s^{-2}}\right)\hat{i}\)
• D. \(-\left(3.5\times10^{13}\,\mathrm{m\,s^{-2}}\right)\hat{i}\)
• A. \(9.0\times10^{-9}\,\text{s}\)
• B. \(1.67\times10^{-8}\,\text{s}\)
• C. \(1.67\times10^{-9}\,\text{s}\)
• D. \(2.17\times10^{-9}\,\text{s}\)
• 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 B

Step 1: Determine the magnitude of electric field between parallel plates. For parallel plate arrangement, \[ E=\frac{V}{d} \] where \[ V=200\,\text{V} \] and \[ d=0.01\,\text{m}. \] Hence, \[ E=\frac{200}{0.01} =2\times10^4\,\text{V m}^{-1}. \]

Step 2: Determine the direction of electric field. Electric field always points from the positive plate towards the negative plate. From the figure, the upper plate is connected to the positive terminal and the lower plate to the negative terminal. Hence field is directed downward. Since positive \(z\)-axis is represented by \(\hat{k}\) and downward direction corresponds to \[ -\hat{k}, \] therefore \[ \boxed{ \vec E=-2\times10^4\,\hat{k}\,\text{V m}^{-1} } \]

Answer 2

Step 1: Calculate force on the electron. Force on a charge is \[ \vec F=q\vec E. \] For an electron, \[ q=-e=-1.6\times10^{-19}\,\text{C}. \] Therefore, \[ \vec F=(-e)(-2\times10^4\hat{k}) \] \[ \vec F = 3.2\times10^{-15}\hat{k}\,\text N. \] Thus force acts along \(+\hat{k}\).

Step 2: Calculate acceleration. Using Newton's second law, \[ a=\frac{F}{m}. \] Mass of electron, \[ m=9.1\times10^{-31}\,\text{kg}. \] Hence, \[ a = \frac{3.2\times10^{-15}} {9.1\times10^{-31}} \] \[ a = 3.52\times10^{15}\,\text{m s}^{-2}. \] Therefore, \[ \boxed{ \vec a= (3.5\times10^{15})\hat{k}\,\text{m s}^{-2} } \]

Answer 3

Step 1: Use horizontal motion. The electron enters with horizontal speed \[ u_x=3\times10^7\,\text{m s}^{-1}. \] Length of plates: \[ L=0.05\,\text m. \] Since electric force acts vertically, horizontal velocity remains constant.

Step 2: Calculate time of travel. \[ t=\frac{L}{u_x} \] \[ t= \frac{0.05} {3\times10^7} \] \[ t= 1.67\times10^{-9}\,\text s. \] Hence, \[ \boxed{ t=1.67\times10^{-9}\,\text s } \]

Answer 4

Step 1: Use vertical motion equation. Initial vertical velocity: \[ u_z=0. \] Vertical acceleration: \[ a=3.5\times10^{15}\,\text{m s}^{-2}. \] Time obtained in part (a): \[ t=1.67\times10^{-9}\,\text s. \]

Step 2: Calculate vertical displacement. \[ s=\frac12 at^2 \] \[ s= \frac12 (3.5\times10^{15}) (1.67\times10^{-9})^2. \] \[ s= 4.88\times10^{-3}\,\text m. \] \[ s=4.88\,\text{mm}. \] Thus, \[ \boxed{ s\approx4.9\,\text{mm} } \]

Answer 5

Step 1: Determine direction of force. Electric field is downward: \[ \vec E=-2\times10^4\hat{k}. \] Since electron has negative charge, \[ \vec F=q\vec E. \] Therefore force acts upward.

Step 2: Determine the nature of motion. The electron has
• constant horizontal velocity,
• uniform upward acceleration. This combination produces a parabolic trajectory.

Step 3: Identify the correct curve. The electron bends upward while moving forward. Among the given curves, path \(b\) represents an upward-opening parabola. Hence, \[ \boxed{\text{Path } b} \] is correct.