Question

In: Physics

In what direction and through what potential difference Vb-Va does the particle move?

A particle with charge 3.20×1019 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.

Part A:
The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x-axis, increasing its kinetic energy by 1.60×1019J. 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 = 0.500 V.
b. The particle moves to the left through a potential difference of Vb-Va = -0.500 V
c. The particle moves to the right through a potential difference of Vb-Va = 0.500 V.
d. The particle moves to the right through a potential difference of Vb-Va = -0.500 V.
e. The particle moves to the left through a potential difference of Vb-Va = 5.00 V.
f. The particle moves to the right through a potential difference of Vb-Va = -5.00 V.

Part B:
If the particle moves from point b to point c in the y-direction, what is the change in its potential energy, Vc-Vb?

a. + 1.60×1019J
b. - 1.60×1019J
c. 0

Solutions

Expert Solution

Concepts and reason

The concepts used to solve the problems are potential energy. Use the kinetic energy and charge to find the direction and potential difference of the particle. Then, use the moving direction of the particle to find the change in potential energy.

Fundamentals

The mathematical expression for the potential energy is as follows:

\(P . E=q(V)\)

Here, the charge on the particle is \(q\), the potential difference is \(P . E\), and the potential difference is \(V\).

 

(A) Write the mathematical expression for the potential energy at point \(a\). \(P . E=q\left(V_{a}\right) \ldots \ldots\) (1)

Here, the charge on the particle is \(q\), and the potential difference at point \(a\) is \(V_{a}\).

Write the mathematical expression for the potential energy at point \(b\). \(P . E=q\left(V_{b}\right) \ldots \ldots(2)\)

Here, the charge on the particle is \(q\), and the potential difference at point \(b\) is \(V_{b}\). Equate the Equation (1) and (2). \(\Delta P . E=q\left|\left(V_{b}-V_{a}\right)\right| \ldots \ldots\) (3)

Here, the change in the potential energy is \(\Delta P . E\). Since the change in kinetic energy is equal to the change in potential energy. Hence Equation (3) is rewritten in terms of kinetic energy \(\Delta K . E=q\left|\left(V_{b}-V_{a}\right)\right|\)

Here, the change in kinetic energy is \(\Delta K . E\). Substitute \(1.6 \times 10^{-19} \mathrm{~J}\) for \(\Delta K . E\) and \(3.20 \times 10^{-19} \mathrm{C}\) for \(q\)

\(1.6 \times 10^{-19}=\left(3.20 \times 10^{-19}\right)\left|\left(V_{b}-V_{a}\right)\right|\)

\(\left|\left(V_{b}-V_{a}\right)\right|=\frac{1.6 \times 10^{-19}}{3.20 \times 10^{-19}}\)

\(\left|\left(V_{b}-V_{a}\right)\right|=0.5 \mathrm{~V}\)

 

Since the potential increases towards positive \(x\) and the electric field are negative \(x\), the particle moves to the left side.

 

Firstly, the corresponding values are substituted in the expression of the kinetic energy to calculate the potential difference.

(B) The incorrect expression is,

- a.) \(+1.60 \times 10^{19} \mathrm{~J}\)

- b.) \(-1.60 \times 10^{19} \mathrm{~J}\)

The change in the potential energy occurs only on \(x\) -axis. And the particle is moving along \(y\) -axis from point \(b\) to point \(c\). There is no change in potential energy on the \(\mathrm{y}\) - axis. Hence, the change in potential energy is zero. Mathematically it can be expressed as follows:

\(V_{c}-V_{b}=0 \mathrm{~V}\)

Hence the correct option is,

- c.) 0

First of all, determine the axis on which the potential energy changes. If the particle is moving on the same axis, then calculate the change in potential. If the change in potential occurs on the other axis, then the value of the change in potential energy is zero.


Part A The particle moves to the left through a potential difference of \(V_{b}-V_{a}=-\mathbf{0 . 5} \mathbf{V}\).

Part B The change in potential energy is zero.

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