Please solve the question carefully and include given, required and solution. So

Question

Please solve the question carefully and include given, required and solution. Solve clearly and neatly.

2) Two parallel plates 2 cm apart form a parallel plate capacitor, and the magnitude of the electric field between the plates is 1.3 x 104 N/C. An electron is given a horizontal velocity of 2.0 x 104 m/s, and is launched at a height halfway between the plates (at a height 1 cm from the bottom plate). After launch the electron is deflected upwards towards the top plate. For what amount of time in seconds does the electron travel from launch until contact with the top plate? What is the direction of the electric field?

Explanation / Answer

Here , the direction of electric field is downwards as the electron is going upwards

Now ,for the acceleration of the electron

a = 1.602 *10^-19 * 1.3 *10^4/(9.11 *10^-31)

a = 2.286 *10^15 m/s^2

let the time taken is t

d = 0.5 * a * t^2

0.01 = 0.5 * 2.286 *10^15 ( t^2 )

solving for t

t = 2.96 *10^-9 s

the time taken for the electron to make contact is 2.96 *10^-9 s

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---