Two parallel plates are separated by 0.1 mm. a 10 v potential difference is maintained between those plates. if a proton is released from the positive plate, calculate the kinetic energy of the proton when it reaches the negative plate.

K.E = 1.6 x 10⁻¹⁸ J

Explanation:

given,                                            

thickness between two plate = t = 0.1 mm            

Voltage  difference between two plate = 10 V          

charge of proton = q = 1.6 x 10⁻¹⁹ C                  

When the charge moves from positive to negative the potential energy reduces to kinetic energy

K.E = Δ PE                                          

K.E = q Δ V                                                

K.E = 1.6 x 10⁻¹⁹  x 10                            

K.E = 1.6 x 10⁻¹⁸ J                                    

so, the kinetic energy of the proton when it reaches negative plate is equal to K.E = 1.6 x 10⁻¹⁸ J

c

explanation:

i believe the answer is c and this is only worth 10 points you might wanna change it

i think its