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Chapter 9: Problem 33

Two oppositely charged parallel plates are held apart at distance of \(10 \mathrm{~cm}\). The potential difference between them is maintained at 50 volts. The electric field between them is (A) \(5 \mathrm{~V} / \mathrm{m}\), down (B) \(50 \mathrm{~V} / \mathrm{m}\), up (C) \(500 \mathrm{~V} / \mathrm{m}\), down (D) \(500 \mathrm{~V} / \mathrm{m}\), up (E) \(1 / 50 \mathrm{~V} / \mathrm{m}\), down

Short Answer

Expert verified
Answer: The electric field between the plates is 500 V/m.

Step by step solution

01

In this scenario, the relationship between electric field (E), potential difference (V), and distance (d) can be described by the following formula: \(E = \frac{V}{d}\) Where E is the electric field, V is the potential difference, and d is the distance between the two plates. #Step 2: Convert the distance to meters#

The distance between the plates is given in centimeters. To properly use the formula, we need the distance in meters. Convert 10 cm to meters: \(d = 10 \mathrm{~cm} \times \frac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.1 \mathrm{~m}\) #Step 3: Calculate the electric field#
02

Using the given potential difference of 50 V and the converted distance of 0.1 meters, apply the formula to find the electric field: \(E = \frac{V}{d} = \frac{50 \mathrm{~V}}{0.1 \mathrm{~m}} = 500 \mathrm{~V} / \mathrm{m}\) #Step 4: Determine the direction of the electric field#

Since the plates are oppositely charged, we can deduce that the electric field will be directed from the positively charged plate to the negatively charged plate. The exercise does not specifically state which plate is positively or negatively charged. Therefore, it is impossible to determine whether the direction of the electric field is up or down. Using the calculated electric field, the closest option in the list given is: (C) \(500 \mathrm{~V} / \mathrm{m}\), down However, without the information on the charge of the plates, we cannot confirm whether the direction is indeed down or up.

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Key Concepts

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

Potential Difference
The concept of potential difference plays a crucial role in understanding electric fields and circuits. In simple terms, it's the difference in electric potential between two points. The potential difference, often referred to as voltage, is what drives electric charges to move from one point to another. It's measured in volts (V).
  • Technically, it's the work done to move a unit charge from one point to another against an electric field.
  • A greater potential difference implies a stronger ability to push charges through a conductor.
In the context of the exercise, the potential difference of 50 volts between the parallel plates indicates that energy is required to move charges between these plates. Understanding this energy difference helps us calculate the electric field, which describes how the potential change occurs over the distance between the plates.
Parallel Plates
Parallel plates in the context of electric fields are a fundamental setup used in physics to study uniform fields. These plates are charged oppositely and create a uniform electric field between them when a potential difference is applied. This setup is often idealized to model simple electric principles.
  • The uniformity of the field means that the electric field strength is constant anywhere between the plates.
  • The distance between the plates affects the strength of the electric field, as seen in the formula for calculating it.
In our exercise, the plates are 10 cm apart, a distance that needs converting to meters for calculations, making it 0.1 m. This distance along with the potential difference helps determine the strength of the electric field, which remains uniform across the space between the plates.
Charge Direction
Determining the direction of an electric field between parallel plates involves understanding the flow of electric charges. Electric fields are vector quantities, which means they have both magnitude and direction.
  • Conventionally, the electric field direction is considered to go from the positive to the negative plate.
  • So, if the top plate is positive and the bottom is negative, the field direction would be downwards.
  • However, the actual physical direction in problems can vary based on the specific charge placements.
In the exercise, it is suggested that the direction of the electric field can't be precisely assigned as up or down without knowing which plate is positively charged. What remains constant, however, is that the field is directed from positive to negative.

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Most popular questions from this chapter

The below fields represent electric or magnetic fields associated with a long current-carrying wire, a single positive charge, two oppositely charged particles, two positive charges, and the field between two oppositely charged parallel sheets. (A) A, D, C, B, E (B) E, A, D, C, B (C) E, D, A, B, C (D) D, E, C, B, A (E) C, B, A, E, D

A test charge \(+q\) and a test charge \(-q\) are released midway between the two plates. Let the voltage of the top plate be \(V\) and the voltage of the bottom plate be 0 . The distance between the two plates is \(d\). Which of the following statements about the test charges are true? i. Both charges have gained a kinetic energy of \(|q| E d / 2\) when they hit the plates. ii. Charge \(+q\) is initially at a positive voltage and charge \(-q\) is initially at a negative voltage. iii. Charge \(+q\) initially has a positive potential energy and charge \(-q\) initially has a negative potential energy. iv. Both charges lose potential energy. (A) i only (B) i and ii (C) i and iii (D) iii and iv (E) i, iii and iv

Until recently, TV sets consisted of what was called a cathode-ray tube. Electrons were boiled off a light bulb filament and passed through plates that deflected the particles; the electron beam went on to strike the TV screen itself and lit up a pixel of phosphor. The beam could be deflected both up and down and sideways (only one direction shown). By deflecting the electron beam and scanning it across the screen a picture was built up. Consider the simplified TV below. An electron beam is shot along the centerline between the two deflector plates. The top plate is held by a power supply (not shown) at a voltage \(V=\) \(+1000\) volts above the bottom plate. The separation between the plates is \(d=4 \mathrm{~cm}\). The length of the plates is \(\ell=8 \mathrm{~cm}\). The distance \(L\) from the end of the deflector plate to the TV screen is \(30 \mathrm{~cm}\). The electrons enter the region between the deflector plates with a velocity \(v=4 \times 10^7 \mathrm{~m} / \mathrm{s}\). a) In which direction is the electric field between the deflector plates? b) In which direction are the electrons deflected? c) What is the force on an electron? (Neglect gravity.) d) What is the acceleration on the electron? e) What is the total deflection of the electron as it hits the TV screen? f) What size magnetic field and in what direction would you need to place between the deflector plates in order to prevent the electrons from being deflected?

Instead of a meter, a device (such as a battery) is attached to the coil that can generate a current. When the device is turned on the bar magnet will (A) remain at rest. (B) be attracted to the coil. (C) be repelled from the coil. (D) shoot through the coil. (E) (B) or (C) depending on which way the current is flowing in the coil

Two identical point charges \(q_1\) and \(q_2\) are at a distance \(r\) apart. If the size of \(q_1\) is doubled and the distance between them tripled, the strength of the electrical force between them (A) goes up by a factor of 3 . (B) goes down by a factor of 3 . (C) goes down by a factor of 9 . (D) goes down by a factor of \(2 / 3\). (E) goes down by a factor of \(2 / 9\)
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