A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 2.0 cmaway, in a time$1.5\times {10}^{-8}s.$. (a) What is the speed of the electron as it strikes the second plate? (b) What is the magnitude of the electric field?

• a)Initial velocity of the electron,$V_I=0m/s$
• b)Distance between two plates,$X=0.02m$

c) Time interval,$t=1.5\times {10}^{-8}S$

The magnitude of the electrostatic force on a point charge of magnitude q is given by F = qE, where E is the magnitude of the electric field at the location of the particle. And the concept of kinematics has been used.

Formulae:

The force due to Newton’s second law, F= ma (i)

The electrostatic force on a body, F= qE (ii)

The distance travelled between two plates,$∆x=V_{avg}torvt/2$(iii)

The second equation of kinematic motion,$∆x=(1/2)a{t}^2(\sin ce{v}_l=0m/s)$(iv)

Using equation (iii), the speed of the electron between the two plates is given as:

$$\begin{gathered} V=\frac{2\left(2.0\times {10}^{-2}m\right)}{1.5\times {10}^{-8}s} \\ =2.7\times {10}^{6}m/s \end{gathered}$$

Hence, the speed of the electron is $2.7\times {10}^{6}m/s$.

From equations (i), (ii) and (iv), the magnitude of the electric field can be given as:

$$\begin{gathered} E=\frac{ma}{e} \\ =\frac{2∆xm}{e{t}^2} \\ =\frac{2\left(2.0\times {10}^{-2}m\right)\left(9.11\times {10}^{-31}kg\right)}{\left(1.60\times {10}^{-19}C\right){\left(1.5\times {10}^{-8}s\right)}^2} \\ =1.0\times {10}^{3}N/C \end{gathered}$$

Hence, the value of the electric field is$1.0\times {10}^{3}N/C$.