Problem 18 - College Physics

Question

A uniform electric field has magnitude \(E\) and is directed in the negative \(x\) -direction. The potential difference between point \(a\) (at \(x=0.60 \mathrm{m} )\) and point \(b\) (at \(x=0.90 \mathrm{m} )\) is 240 \(\mathrm{V}\) . (a) Which point, \(a\) or \(b\) , is at the higher potential? (b) Calculate the value of \(E (\mathrm{c})\) A negative point charge \(q=-0.200 \mu \mathrm{C}\) is moved from \(b\) to \(a\) . Calculate the work done on the point charge by the electric field.

Step-by-Step Solution

Answer

(a) Point a is at a higher potential. (b) E = 800 V/m. (c) Work done = -48 µJ.

Step 1: Analyzing Potential Difference and Electric Field Direction

Since the electric field is uniform and directed in the negative \(x\)-direction, it points from point \(b\) to \(a\). The potential difference \(V_{ab} = V_a - V_b = 240\; \mathrm{V}\). This means \(a\) is at a higher potential than \(b\) because the potential decreases in the direction of the electric field.

Step 2: Calculating Electric Field Magnitude

The potential difference \(V_{ab} = E \cdot d\), where \(d\) is the distance between the two points. Since \(d = 0.90 - 0.60 = 0.30\; \mathrm{m}\), we can solve for \(E\) as follows:\[E = \frac{V_{ab}}{d} = \frac{240\; \mathrm{V}}{0.30\; \mathrm{m}} = 800\; \mathrm{V/m}.\]

Step 3: Calculating Work Done on the Charge

The work done by the electric field when moving a charge \(q\) from point \(b\) to point \(a\) is given by \(W = q \cdot V_{ab}\). Here, \(q = -0.200 \times 10^{-6}\; \mathrm{C}\) and \(V_{ab} = 240\; \mathrm{V}\):\[W = (-0.200 \times 10^{-6}) \cdot 240 = -48 \times 10^{-6}\; \mathrm{J} = -48\; \mu\mathrm{J}.\] The negative sign indicates that the work is done on the electric field by the charge.

Key Concepts

Potential Difference

The potential difference in an electric field is a measure of how much potential energy is available to move charged particles between two points. It represents the energy difference per unit charge, which is usually expressed in volts (V). In the scenario where a uniform electric field is present, the potential difference is directly related to the work required to move a charge between two points.

In this case, point "a" is at a higher electrical potential than point "b." Since the electric field directs from higher to lower potential, it flows from "a" to "b." This is aligned with the given potential difference of 240 V between these points, confirming that more energy per unit charge exists at "a" compared to "b." Meaning, if you place a positive charge at point "a," it would naturally move towards point "b" under the influence of the electric field, releasing energy as it goes.

Work Done

In an electric field, work is done whenever a charge moves from one point to another. This is because it takes energy to move the charge, especially if it goes against the direction of the electric field. The formula for work done in this context is given by:

• \( W = q \cdot V_{ab}, \text{ where } q \text{ is the charge and } V_{ab} \text{ is the potential difference}. \)

In the given problem, a negative charge of value \(-0.200 \mu C\) is moved from point "b" to point "a." The potential difference of 240 V tells us how much energy per unit charge is involved in this movement. When you calculate using the formula, the work done is \(-48 \mu J\), demonstrating that the work is energy given to the electric field. The negative result indicates that energy is needed to move the charge "against" its natural path, confirming movement from a region of lower potential to higher potential.

Uniform Electric Field

A uniform electric field is one where the field strength is constant in magnitude and direction throughout the space. This means that the force experienced by a charged particle within the field does not change as it moves along the field lines.

In this exercise, the electric field is noted to be uniform and directed in the negative x-direction. This means that if you were to plot the field lines, they would be evenly spaced and all pointing straight along the negative x-axis. The uniform aspect simplifies calculations as the potential difference can be directly related to the distance between the points and the field strength. It permits the formula:

• \( V_{ab} = E \cdot d, \text{ where } E \text{ is the electric field strength and } d \text{ is the distance between the points}. \)

Using the given values, this leads us to calculate that the field strength is 800 V/m, meaning every meter along this field sees a change in potential of 800 volts.

Electric Potential

Electric potential is a scalar quantity that represents the potential energy per unit charge at a point in an electric field. It is measured in volts and provides crucial insights into the behavior of charged particles within the field. Higher potential means more energy stored per charge, guiding the direction and movement of charges.

For the exercise at hand, point "a" at 0.60 m holds a higher electric potential compared to point "b" at 0.90 m. Electric potential follows the field lines, decreasing in the direction of the field if we move from positive to negative region in terms of charge. This aligns with the electric field being directed towards the negative x-direction. In any electrical system, charges spontaneously move from regions of higher to lower potential, developing flow or current if allowed. This principle is crucial in electrical circuits and influences how we harness electrical energy in various applications.