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The amount of work done to move a positive point charge \(q\) on an equipotential surface of \(1000 \mathrm{~V}\) relative to that on an equipotential surface of \(10 \mathrm{~V}\) is a) the same. d) dependent on the b) less. distance the charge c) more. moves.

Short Answer

Expert verified
Actually, the correct answer should be: d) dependent on the value of the charge q. The work done depends on the charge q, not on the distance the charge moves. I apologize for the mistake in the initial response.

Step by step solution

01

Identify the given variables and constants.

Here, we are given: Voltage of surface 1, V1 = 1000 V Voltage of surface 2, V2 = 10 V Charge: q
02

Calculate the potential difference between the two surfaces.

To find the potential difference between the two equipotential surfaces, subtract the potential of surface 2 from the potential of surface 1. ∆V = V1 - V2 = 1000 V - 10 V = 990 V
03

Calculate the work done to move the charge between the two surfaces.

Now we can calculate the work done (W) to move the positive point charge q between these two surfaces. W = q∆V Since this formula includes q, and no further information is given about q or the distance moved, the work done will depend on the charge.
04

Identify the answer in the options.

Based on the calculations, the amount of work done to move the positive point charge q between the two equipotential surfaces is dependent on the charge. Thus, the correct answer is: d) dependent on the distance the charge moves.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Potential Difference
To understand the concept of electric potential difference, let's consider a familiar analogy. Imagine you have two water tanks at different heights. The difference in height creates a pressure difference, which could potentially push water from the high tank to the low one if a path is open. Similarly, in the world of electricity, the electric potential difference (often just called voltage) between two points is what 'pushes' electric charges from a point of high potential energy to low potential energy.

Using the formula for potential difference, \( \Delta V = V_1 - V_2 \) where \( V_1 \) and \( V_2 \) are the potentials at two different points, we find that moving a charge between these two points involves a specific change in potential. Going back to our textbook exercise, a positive point charge moving from a spot with 1000 V to one with 10 V experiences a potential difference of 990 V. This difference is what 'encourages' the charge to move, just like water naturally flows downhill.
Work Done by Electric Force
Imagine pushing a ball up a hill. The work you do is equal to the force you apply times the distance you push it. In the electrical world, we define work done by electric force similarly, using the equation \( W = q\Delta V \) where \( W \) is the work done by the electric force, \( q \) is the charge, and \( \Delta V \) is the electric potential difference.

When you're considering work done on a charge in an electric field, it's crucial to realize that if the electric potential doesn't change—as on an equipotential surface—no work is needed to move that charge on that surface. That's because the force the electric field exerts on the charge at every point on that surface does not have a component in the direction of the movement, just like you do no work balancing a book on your head while walking flat ground. But, as soon as there is a potential difference, as with our 990 V from earlier, work comes into play, and it is proportional to both the charge and the electric potential difference.
Positive Point Charge
Exploring the role of a positive point charge in this electrical dance, it helps to first clarify what we mean by a point charge. A point charge is a charged object considered to be dimensionless; it serves as an idealization to simplify the analysis of electric fields. A positive point charge, specifically, has an excess of protons and is drawn toward areas of lower electric potential, or towards negative charges.

In our exercise scenario, the positive point charge is the 'character' moving from one equipotential surface to another. The electric potential difference acts somewhat like a downhill slope for this charge. Since work is done on this charge as it moves between different voltages, and the amount of work depends on the magnitude of the charge itself, it's important to remember that charges in electric fields don't just float around aimlessly—they respond actively to the 'landscape' of electric potential.

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Most popular questions from this chapter

High-voltage power lines are used to transport electricity cross country. These wires are favored resting places for birds. Why don't the birds die when they touch the wires?

A negatively charged particle revolves in a clockwise direction around a positively charged sphere. The work done on the negatively charged particle by the electric field of the sphere is a) positive. b) negative. c) zero.

The electric field, \(\vec{E}(\vec{r}),\) and the electric potential \(V(\vec{r}),\) are calculated from the charge distribution, \(\rho(\vec{r}),\) by integrating Coulomb's Law and then the electric field. In the other direction, the field and the charge distribution are determined from the potential by suitably differentiating. Suppose the electric potential in a large region of space is given by \(V(r)=V_{0} \exp \left(-r^{2} / a^{2}\right),\) where \(V_{0}\) and \(a\) are constants and \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) is the distance from the origin. a) Find the electric field \(\vec{E}(\vec{r})\) in this region. b) Determine the charge density \(\rho(\vec{r})\) in this region, which gives rise to the potential and field. c) Find the total charge in this region. d) Roughly sketch the charge distribution that could give rise to such an electric field.

How much work would be done by an electric field in moving a proton from a point at a potential of \(+180 . \mathrm{V}\) to a point at a potential of \(-60.0 \mathrm{~V} ?\)

A deuterium ion and a tritium ion each have charge \(+e .\) What work is necessary to be done on the deuterium ion in order to bring it within \(10^{-14} \mathrm{~m}\) of the tritium ion? This is the distance within which the two ions can fuse, as a result of strong nuclear interactions that overcome electrostatic repulsion, to produce a helium-5 nucleus. Express the work in electron-volts.

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