Question

A particle with charge 8.00×10^−19 C is placed on the x axis in a region where the electric potential due to other charges increases in the +x direction but does not change in the y or z direction.

The particle, initially at rest, is acted upon only by the electric force and moves from point a to point balong the x axis, increasing its kinetic energy by 1.60×10^−18 J . In what direction and through what potential difference Vb−Va does the particle move?

A. The particle moves to the left through a potential difference of Vb−Va= 2.00 V .
B. The particle moves to the left through a potential difference of Vb−Va= -2.00 V
C. The particle moves to the right through a potential difference of Vb−Va= 2.00 V .
D. The particle moves to the right through a potential difference of Vb−Va= -2.00 V .
E. The particle moves to the left through a potential difference of Vb−Va= 20.0 V .
F. The particle moves to the right through a potential difference of Vb−Va= -20.0 V .

Answer 1

The particle, which is positively charged, moves from higher to lower potential, meaning it moves in the direction of the electric field. Given the increase in kinetic energy and the charge of the particle, we calculate that the particle moves to the right through a potential difference of Vb - Va = 2.00 V.

Step-by-step explanation:

To determine the direction and potential difference through which the particle moves, we can use the conservation of energy. The work done by the electric field on the particle equals the change in its kinetic energy when it is moved from a point a to point b. The electric potential energy of the particle is given by U = qV, where q is the charge of the particle and V is the electric potential.

Since the particle's kinetic energy increases, it means that it has moved from a higher potential to a lower potential (i.e., work is done by the electric field). The change in kinetic energy (ΔKE) is equal to the charge (q) multiplied by the potential difference (ΔV), which is U₂ - U₁ or q(Vb - Va).

Given that ΔKE = 1.60×10⁻¹⁸ J and q = 8.00×10⁻¹⁹ C, we calculate Vb - Va by rearranging the equation ΔKE = q(Vb - Va):

Vb - Va = ΔKE / q = (1.60×10⁻¹⁸ J) / (8.00×10⁻¹⁹ C) = 2.00 V.

Since the charge is positive and moves from higher to lower potential, it moves in the direction of the electric field, which is the +x direction. Therefore, the particle moves to the right through a potential difference of Vb - Va = 2.00 V.