A test charge \(+q\) and a test charge \(-q\) are released midway between the two plates. Let the voltage of the top plate be \(V\) and the voltage of the bottom plate be 0 . The distance between the two plates is \(d\). Which of the following statements about the test charges are true? i. Both charges have gained a kinetic energy of \(|q| E d / 2\) when they hit the plates. ii. Charge \(+q\) is initially at a positive voltage and charge \(-q\) is initially at a negative voltage. iii. Charge \(+q\) initially has a positive potential energy and charge \(-q\) initially has a negative potential energy. iv. Both charges lose potential energy. (A) i only (B) i and ii (C) i and iii (D) iii and iv (E) i, iii and iv

As the voltage of the top plate is V, and the voltage of the bottom plate is 0, the total voltage difference between the two plates is V. The distance between the two plates is d. The electric field (E) between the plates can be calculated using the formula: \(E = \frac{V}{d}\)

Potential energy (PE) can be calculated using the formula: \(PE = qV\) Since the charges are midway between the plates, the potential energy of each charge is: For charge +q: \(PE_+ = +q(\frac{V}{2})\) For charge -q: \(PE_- = -q(\frac{V}{2})\)

i. As both charges move towards the plates (the positive charge towards the bottom plate, and the negative charge towards the top plate), their potential energy is converted into kinetic energy (KE). When both charges reach the plates, their potential energies will have been completely converted into kinetic energies: For charge +q: \(KE_+ = \frac{1}{2}|q|Ed\) For charge -q: \(KE_- = \frac{1}{2}|q|Ed\) The statement is true. ii. Voltage is defined as the potential energy per unit charge, and midway between the plates, both test charges experience half the total voltage, \(V/2\). The positive test charge (+q) will be at a potential of \(+V/2\), and the negative test charge (-q) will be at a potential of \(-V/2\). Therefore, this statement is true. iii. From step 2, we know that the potential energy of the positive test charge (+q) is positive, while the potential energy of the negative test charge (-q) is negative. Thus, this statement is true. iv. When both charges move towards the plates, their potential energies decrease (as calculated in step 2). Consequently, this statement is also true.

All the given statements (i, ii, iii, and iv) are true. Therefore, the correct answer is (E) i, iii, and iv.