Aproton is placed in an electric field of intensity 500 n/c what is the magnitude and direction of the acceleration of this proton due to this field

First, we identify the given values in this item:
Electric field intensity, E = 500 N/C

The charge, q, of proton is determined to be equal to 1.6 x10^-19 C.

Electric force given the electric field intensity and charge, can be calculated through the equation,
                     Fe = qE
Substituting the known values,
                     Fe = (1.6 x 10^-19 C) x (500N)
                      Fe = 8 x10^-17 N

Acceleration is calculated by dividing Fe by the mass. The mass of the proton is equal to 9.11 x 10^-31 kg.
                     a = (8 x 10^-17 N) / (9.11 x 10^-31 kg)
                      a = 8.78 x 10^13 m/s²

Thus, the acceleration is 8.78 x 10^13 m/s² and the direction is in the positive direction.