Final answer:
The electric field strength between the plates, which accelerates the doubly charged ion to an energy of 32.2 keV over a separation distance of 1.83 cm, is approximately 1.039 × 10^6 V/m.
Explnation:
The kinetic energy acquired by the doubly charged ion in an electric field can be calculated using the formula: KE = qV, where KE is the kinetic energy, q is the charge of the ion, and V is the potential difference it moves through.
Given the kinetic energy of the ion as 32.2 keV, and since 1 keV is equal to 1.6 × 10^(-16) J, we convert the kinetic energy to joules: 32.2 × 10^3 × 1.6 × 10^(-16). This calculation provides us with the kinetic energy in joules.
Next, we utilize the equation for the kinetic energy of a charged particle accelerated through a potential difference: KE = qV. Here, q is the charge of the ion and V is the potential difference.
As the ion is accelerated by the electric field between the plates separated by a distance of 1.83 cm, we rearrange the equation to solve for the potential difference V. Knowing the charge of the ion (twice the elementary charge), we can compute the potential difference. Once we have the potential difference, we apply the formula for the electric field strength E = V/d, where E represents the electric field strength and d is the separation distance between the plates (1.83 cm converted to meters).
Substituting the calculated potential difference into the equation for electric field strength, we determine the electric field strength between the plates to be approximately 1.039 × 10^6 V/m. This value signifies the intensity of the electric field that accelerates the doubly charged ion between the parallel conducting plates.