A doubly charged ion is accelerated to an energy of 33.0 keV by the electric field between two parallel conducting plates separated by 1.70 cm. What is the electric field strength between the plates?

To find the electric field strength between the plates, we can use the formula:

E = V/d

where E is the electric field strength, V is the potential difference, and d is the distance between the plates.

In this case, the potential difference (V) is given as 33.0 keV, which can be expressed in joules using the conversion factor: 1 keV = 1.6 x 10^-19 J.

Therefore, V = 33.0 keV * (1.6 x 10^-19 J/1 keV) = 5.28 x 10^-18 J.

The distance between the plates (d) is given as 1.70 cm, which can be converted to meters by dividing by 100: d = 1.70 cm / 100 = 0.017 m.

Now we can substitute the values into the formula to find the electric field strength:

E = V/d = (5.28 x 10^-18 J) / (0.017 m) = 3.11 x 10^-16 N/C.

Therefore, the electric field strength between the plates is 3.11 x 10^-16 N/C.

To find the electric field strength between the plates, we can use the formula:

E = V / d,

where E is the electric field strength, V is the voltage, and d is the distance between the plates.

In this case, we are given the energy of the charged ion as 33.0 keV, which can be converted to voltage using the formula:

V = q * E,

where q is the charge of the ion.

Since the ion is doubly charged, the charge (q) will be twice the elementary charge (e) of an electron.

So, q = 2e,

where e = 1.6 x 10^-19 C (the elementary charge).

Now, we can calculate the voltage (V):

V = (2 * 1.6 x 10^-19 C) * (33.0 x 10^3 eV) / 1 eV,

V = 1.056 x 10^-17 C * 33,000,

V = 3.4848 x 10^-13 C.

Next, we need to find the distance (d) between the plates. The problem states that the plates are separated by 1.70 cm.

Now, we can substitute the values of V and d into the formula for electric field strength:

E = (3.4848 x 10^-13 C) / (1.70 x 10^-2 m),

E = 2.047 x 10^-11 C/m.

Thus, the electric field strength between the plates is 2.047 x 10^-11 C/m.