A Van de Graaff generator is charged so that the magnitude of the electric field at its surface is 3.0 multiplied by 104 N/C.

(a) What is the magnitude of the electric force on a proton released at the surface of the generator?

(b) Find the proton's acceleration at this instant.

To solve this problem, we need to use the equation for electric force (F = qE) and the equation for acceleration (a = F/m), where q is the charge of the proton, E is the electric field, F is the electric force, and m is the mass of the proton.

Let's first find the magnitude of the electric force on the proton (F):

(a) The magnitude of the electric field (E) is given as 3.0 x 10^4 N/C.

The charge of a proton (q) is 1.6 x 10^-19 C.

Using the equation for electric force (F = qE), we can calculate the electric force on the proton as:

F = qE
= (1.6 x 10^-19 C) * (3.0 x 10^4 N/C)
= 4.8 x 10^-15 N

Therefore, the magnitude of the electric force on the proton is 4.8 x 10^-15 N.

Now, let's find the proton's acceleration (a):

(b) The mass of a proton (m) is approximately 1.67 x 10^-27 kg.

Using the equation for acceleration (a = F/m), we can calculate the proton's acceleration as:

a = F/m
= (4.8 x 10^-15 N) / (1.67 x 10^-27 kg)
≈ 2.87 x 10^12 m/s^2

Therefore, the proton's acceleration at this instant is approximately 2.87 x 10^12 m/s^2.