Answer:
15000 V/m
![2.634467618* 10^(15)\ m/s^2](https://img.qammunity.org/2021/formulas/physics/college/gmz7quyqb8fc200v1wwaoqgph3j1p1gheu.png)
Step-by-step explanation:
V = Voltage = 30 V
d = Separation = 2 mm
q = Charge of electron =
![1.6* 10^(-19)\ C](https://img.qammunity.org/2021/formulas/physics/high-school/nec44trxsj2caqi3gb0le5xflruyaun4do.png)
m = Mass of electron =
![9.11* 10^(-31)\ kg](https://img.qammunity.org/2021/formulas/physics/high-school/3pib3fvzuqmvt7snsfwbcgy613zwzdy38c.png)
Electric field is given by
![E=(V)/(d)\\\Rightarrow E=(30)/(2* 10^(-3))\\\Rightarrow E=15000\ V/m](https://img.qammunity.org/2021/formulas/physics/college/ct3htg1y5hk58ntc3kfcl8di14urvh0nmr.png)
The electric field between the plates is 15000 V/m
Acceleration is given by
![a=(qE)/(m)\\\Rightarrow a=(1.6* 10^(-19)* 15000)/(9.11* 10^(-31))\\\Rightarrow a=2.634467618* 10^(15)\ m/s^2](https://img.qammunity.org/2021/formulas/physics/college/rskp2p0jlqxk7pndc0sft2jq6k0vbzxhu6.png)
The acceleration is
![2.634467618* 10^(15)\ m/s^2](https://img.qammunity.org/2021/formulas/physics/college/gmz7quyqb8fc200v1wwaoqgph3j1p1gheu.png)