
How does the voltmeter reading compare to the potential difference across the electrodes? Explain.
If the sliding lead from electrode A were connected at point C along the resistor, would the voltmeter reading be positive, negative, or zero? Explain.
(Hint: Imagine disconnecting the ammeter and evacuated tube from the rest of the circuit, and answering the same question.)
How would you adjust the sliding connection from electrode A in order to make the potential difference across the electrodes

The potential difference reading in the voltmeter.
Voltmeter reading positive, negative or zero.
To adjustment to be made for the Voltmeter reading as positive and negative.
Answer to Problem 1aT
The potential across the electrodes is equal to the potential difference reading in the voltmeter.
Voltmeter reading will be positive.
By reversing the direction of the connection of the variable battery voltmeter can read negative potential.
Explanation of Solution
Introduction:
Photoelectric effect: Electrons from a metal surface are ejected when a light of an appropriate frequency is incident on it.The ejected electrons are called photoelectrons and the whole phenomenon is called photoelectric effect. This effect was explained by The Albert Einstein.
Figure 1 shows the circuit diagram to study the Photoelectric effect (PE).
Figure 1: Set up to study photoelectric effect
A monochromatic is incident on one of the electrodes ‘b’ . Electron bonded to a metal requires a minimum energy just to leave the surface of metal is called binding energy of electron, also known as work function of the electron and denoted as
Current is flowing from positive to negative terminal of the battery (shown as red arrows in Figure 1).Therefore, the voltmeter will show the potential across electrode positive as electrons are flowing from high potential to lower potential.
Figure 2 shows the circuit diagram to study the Photoelectric effect (PE). In order to obtain the positive potentialdifference, read by voltmeter, figure 2(1) is applicable, as the electrode ‘b’ exposed to light is connected to negative terminal and other electrode ‘a’to the positive terminal of the battery. To make it more positive, one need to increase the potential of the variable battery that will make the electrode ‘a’more positive, hence will result in more positive potential across the electrodes.
Figure 2: Set up to study photoelectric effect
In order to obtain the negative potential difference,read by voltmeter, figure 2(2) is applicable, as the electrode ‘b’ exposed to light is connected to positive terminal and other electrode ‘a’to the negative terminal of the battery.
Conclusion:
The potential across the electrodes is equal to the potential difference reading in the voltmeter.Voltmeter reading will be positive.By reversing the direction of the connection of the variable battery voltmeter can read negative potential.
Want to see more full solutions like this?
Chapter 14 Solutions
Tutorials in Introductory Physics
Additional Science Textbook Solutions
Cosmic Perspective Fundamentals
Anatomy & Physiology (6th Edition)
Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)
Chemistry: Structure and Properties (2nd Edition)
Microbiology: An Introduction
Campbell Biology (11th Edition)
- You are standing a distance x = 1.75 m away from this mirror. The object you are looking at is y = 0.29 m from the mirror. The angle of incidence is θ = 30°. What is the exact distance from you to the image?arrow_forwardFor each of the actions depicted below, a magnet and/or metal loop moves with velocity v→ (v→ is constant and has the same magnitude in all parts). Determine whether a current is induced in the metal loop. If so, indicate the direction of the current in the loop, either clockwise or counterclockwise when seen from the right of the loop. The axis of the magnet is lined up with the center of the loop. For the action depicted in (Figure 5), indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop). I know that the current is clockwise, I just dont understand why. Please fully explain why it's clockwise, Thank youarrow_forwardA planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).arrow_forward
- What are the expected readings of the ammeter and voltmeter for the circuit in the figure below? (R = 5.60 Ω, ΔV = 6.30 V) ammeter I =arrow_forwardsimple diagram to illustrate the setup for each law- coulombs law and biot savart lawarrow_forwardA circular coil with 100 turns and a radius of 0.05 m is placed in a magnetic field that changes at auniform rate from 0.2 T to 0.8 T in 0.1 seconds. The plane of the coil is perpendicular to the field.• Calculate the induced electric field in the coil.• Calculate the current density in the coil given its conductivity σ.arrow_forward
- An L-C circuit has an inductance of 0.410 H and a capacitance of 0.250 nF . During the current oscillations, the maximum current in the inductor is 1.80 A . What is the maximum energy Emax stored in the capacitor at any time during the current oscillations? How many times per second does the capacitor contain the amount of energy found in part A? Please show all steps.arrow_forwardA long, straight wire carries a current of 10 A along what we’ll define to the be x-axis. A square loopin the x-y plane with side length 0.1 m is placed near the wire such that its closest side is parallel tothe wire and 0.05 m away.• Calculate the magnetic flux through the loop using Ampere’s law.arrow_forwardDescribe the motion of a charged particle entering a uniform magnetic field at an angle to the fieldlines. Include a diagram showing the velocity vector, magnetic field lines, and the path of the particle.arrow_forward
- Discuss the differences between the Biot-Savart law and Coulomb’s law in terms of their applicationsand the physical quantities they describe.arrow_forwardExplain why Ampere’s law can be used to find the magnetic field inside a solenoid but not outside.arrow_forward3. An Atwood machine consists of two masses, mA and m B, which are connected by an inelastic cord of negligible mass that passes over a pulley. If the pulley has radius RO and moment of inertia I about its axle, determine the acceleration of the masses mA and m B, and compare to the situation where the moment of inertia of the pulley is ignored. Ignore friction at the axle O. Use angular momentum and torque in this solutionarrow_forward
- Glencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-HillCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax College
- University Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice UniversityPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning





