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Two very large metal parallel plates that are \(25 \mathrm{~cm}\) apart, oriented perpendicular to a sheet of paper, are connected across the terminals of a \(50.0 \mathrm{~V}\) battery. (a) Draw to scale the lines where the equipotential surfaces due to these plates intersect the paper. limit your drawing to the region between the plates, avoiding their edges, and draw the lines for surfaces that are \(10.0 \mathrm{~V}\) apart, starting at the lowpotential plate. (b) These surfaces are separated equally in potential. Are they also separated equally in distance? (c) In words, describe the shape and orientation of the surfaces you just found.

Short Answer

Expert verified
Equipotential lines are horizontal, spaced every 5 cm. They are equally spaced in both potential and distance. They are flat planes parallel to the plates.

Step by step solution

01

Determine Equipotential Boundaries

The plates are connected to a 50.0 V battery and are 25 cm apart. The potential drop between them is 50.0 V, from the high-potential plate at 50.0 V to the low-potential plate at 0 V. The equipotential surfaces are the locations where the potential remains constant.
02

Calculate Equipotential Increments

The problem asks for equipotential lines at every 10.0 V. Thus, these equipotential surfaces will occur at 0 V, 10 V, 20 V, 30 V, 40 V, and 50 V since the total potential is 50 V.
03

Calculate Distance between Equipotential Surfaces

The distance between the plates is 25 cm, covering a 50.0 V change. Since we want equipotentials every 10 V, we will have 5 boundaries (0-10 V, 10-20 V, etc.). The distance per equipotential change is: \[ \frac{25\, \text{cm}}{5 \text{ equipotentials}} = 5\, \text{cm} \] for each 10 V change.
04

Draw Equipotential Lines

To draw the equipotential lines, start at the low potential plate. Draw horizontal lines at intervals of 5 cm vertically apart on the paper (assuming the paper represents a cross-section between plates), which correlate with 0 V, 10 V, 20 V, 30 V, 40 V, and 50 V potentials.
05

Analyze Separation Equality (Part b)

As the electric field is uniform between large parallel plates, both the potential difference and distance are uniform. Hence, equipotential surfaces are equally spaced, both in terms of potential and distance.
06

Describe Shape and Orientation (Part c)

Equipotential surfaces between large parallel plates are uniform plane surfaces parallel to the plates themselves. They are evenly spaced and maintain this uniformity regardless of distance from the plates.

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Key Concepts

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

Equipotential Surfaces
Equipotential surfaces represent points in an electric field that share the same electric potential. This implies that if you were to move along an equipotential surface, no work would be required to maintain your movement. Between two parallel plates, these surfaces are planar and align parallel to the plates. Surrounding high and low potential plates, these surfaces can be visualized as cutting through space evenly, maintaining team alignment and distance.

In the case of our exercise, we have electric potential changes of every 10 V. The surfaces of 0 V, 10 V, 20 V, 30 V, 40 V, and 50 V are all equally distant from each other in the region between the plates. This is explained by the uniformity of the electric field in this scenario.
Parallel Plates
When dealing with electric fields, parallel plates serve as a classic configuration to study. Parallel plates refer to two large conductive plates set parallel to each other, often connected to a voltage source to create an electric field. The simplicity of this arrangement provides a uniform electric field between the plates.

Very large plates ensure edge effects are minimized. This stability offers predictable behavior of charged particles and equipotential surfaces between them. In this exercise, the plates are separated by 25 cm and maintain a potential difference of 50 V. When plates are sufficiently large, the electric field they produce is considered uniform across the central region between them.
Electric Field Uniformity
Electric field uniformity is a key feature in this exercise. In a uniform electric field, the potential difference per unit distance remains constant, making calculations straightforward and outcomes predictable.

For parallel plates connected to a voltage, the electric field strength, denoted as E, is consistent. It can be calculated using the formula: \[ E = \frac{V}{d} \] where V is the voltage difference and d is the distance between the plates. Here, for our 50 V potential difference across 25 cm, the electric field’s strength remains steady, at 2 V/cm. This uniform electric field means that each equipotential surface is equidistant in voltage and actual space between the plates, as seen in our regular 5 cm spacing for every 10 V increment.
Voltage Difference
Voltage difference, or potential difference, is the measure of electrical potential energy between two points in a circuit. It drives the current from a point of high potential to a point of lower potential. In the battery-powered system between our parallel plates, a 50 V difference is sustained.

This voltage creates an electric field that drives the formation and positioning of equipotential surfaces. Between the plates, this difference is divided into equal increments (10 V each) for visualization on equipotential surfaces. Known as homogeneous spacing, its effects preserve consistent behavior and spacing of the surfaces as particles travel between high and low potential zones. The 50 V difference helps define both the intensity of the electric field and the spacing of the equipotential lines, showcasing the fundamental principle of voltage driving electric fields.

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Most popular questions from this chapter

A parallel-plate air capacitor has a capacitance of \(500.0 \mathrm{pF}\) and a charge of magnitude \(0.200 \mu \mathrm{C}\) on each plate. The plates are \(0.600 \mathrm{~mm}\) apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electric-field magnitude between the plates? (d) What is the surface charge density on each plate?

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 \(E\). (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.

The plates of a parallel-plate capacitor are \(2.50 \mathrm{~mm}\) apart, and each carries a charge of magnitude \(80.0 \mathrm{nC}\). The plates are in vacuum. The electric field between the plates has a magnitude of \(4.00 \times 10^{6} \mathrm{~V} / \mathrm{m}\). (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the capacitance?

X-ray tube. An X-ray tube is an evacuated glass tube that produces electrons at one end and then accelerates them to very high speeds by the time they reach the other end. The acceleration is accomplished using an electric field. The high-speed electrons hit a metal target at the other end, and the violence of the collision converts their kinetic energy into high-energy light rays, commonly known as X-rays. (a) Through what potential difference should electrons be accelerated so that their speed is \(1.0 \%\) of the speed of light when they hit the target? (b) What potential difference would be needed to give protons the same kinetic energy as the electrons? (c) What speed would this potential difference give to protons? Express your answer in \(\mathrm{m} / \mathrm{s}\) and as a percent of the speed of light.

How far from a \(-7.20 \mu \mathrm{C}\) point charge must a \(+2.30 \mu \mathrm{C}\) point charge be placed in order for the electric potential energy of the pair of charges to be \(-0.400 \mathrm{~J}\) ? (Take the energy to be zero when the charges are infinitely far apart.)

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