Question

A doubly charged ion is accelerated to an energy of 32.0 keV by the electric field between two parallel conducting plates separated by 2.00 cm. What is the electric field strength between the plates?

A) (800 V/m)
B) (1.0 times 10^6 V/m)
C) (4.0 times 10^6 V/m)
D) (1.6 times 10^6 V/m)

Answer 1

To find the electric field strength, use the formula E = V/d, by calculating the potential difference V from the given energy and charge of the ion, and then dividing it by the separation distance d. The correct electric field strength is 1.6 × 10^6 V/m.

Step-by-step explanation:

The electric field strength between two parallel conducting plates can be found by using the formula E = V/d, where V is the potential difference and d is the separation distance between the plates. We are given that the doubly charged ion is accelerated to an energy of 32.0 keV.

Since 1 eV = 1.602 × 10-19 joules and the ion has two charges (2e), the potential difference V that the ion experiences is given by V = energy/charge = 32.0 keV × 103 eV/keV × 1.602 × 10-19 J/eV / (2 × 1.602 × 10-19 C). This simplifies to V = 32.0 × 103 volts. Now, placing this value into the formula for electric field strength and using the plate separation of 2.00 cm (0.02 m), we get E = V/d = 32.0 × 103 V / 0.02 m, which results in E = 1.6 × 106 V/m. Therefore, the correct answer is 1.6 × 106 V/m, which corresponds to option D).