VDU_Physics_3_008.htm

# ## Solved University-Physics Problems

Physics 3:  Introduction to Electricity & Magnetism

## <<<  Electric Potential  >>>

Javier Montenegro Joo

jmj@VirtualDynamicsSoft.com

(1) Two parallel plates of length L and separation d, (L>>d) have opposite charges so that there is a vertical electric field and a potential difference Vo, between them.   A particle of mass m, charge Q and horizontal velocity   Vox   enters the region between the plates, at the level of the positive plate. Neglecting the effect of the acceleration of the gravity, determine the vertical position of the particle, when it has traveled a horizontal distance equal to half the separation between the plates.

Solution.-

The electric field between the parallel plates accelerates upwards the positively charged particle. As a consequence of the combined effects of the horizontal velocity of the particle and the vertical acceleration, the particle displaces along an upward parabolic trajectory between the plates. (2) Between two vertical charged parallel plates exists a uniform electric field. Inside the field there is a particle of mass m and charge Q, displacing with velocity  vo   Determine (a) The potential difference to be applied in order to reduce in   20%   the speed of the particle (b) The change in the kinetic energy of the particle.

Solution

Any work made on a particle produces in a change in its kinetic energy.

A force applied along a distance ‘d’ is work:      W = F d   and work is energy.

Reducing the speed of a particle is equivalent to decelerate it and to reduce its kinetic energy. In order to decelerate a positive electric charge, a negative potential must be applied. Doppler Effect Simulation Module in the Physics Virtual Lab (PVL)

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(3) Consider a spherical distribution of electric charge expressed by For an outer radial distance r, determine the electric field and the electric potential.

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(4) A spherical shell made of conductive material, with radiuses ‘a’ and ‘b’, (a < b), has an electrical charge density of  10 Qo/3.    For points outside the shell, compute (a) Electric field (b) electric potential.

Solution.-

Since the shell is made of conductive material, the electric charge distributes only on its surface, hence the mentioned density is superficial.  At the time of calculating the electric potential it must be borne in mind that very far from the shell (at the infinite), the effect of its electrical charge (its potential) is null. (5) Two very long cylindrical and concentric shells, made of metal, with radiuses r1 and r2, (r1 < r2), possess charge densities Lambda and –Lambda, respectively. It is observed that at a distance r0, such that r0 > r2, the electric potential is zero. Determine the potential difference between the cylindrical shells.

Solution.-

Note that in this system of two oppositely charged concentric cylinders the total electric charge is null, because positive and negative charges of equal magnitude cancel each other. In the region inside r1 there is no charge. Then there is no electric field, neither in the outer region of the cylinders, nor in their inner region. Electric field exists only in the region between both cylinders and, this field is due only to the charge in the inner shell. The charge of the outer cylinder does not contribute to the electric field between both shells. Algorithmic Art by Javier Montenegro Joo To appreciate astonishing algorithmic art images by JMJ visit:

(6) Four electric charges,   -3Qo ,   6Qo ,  -9Qo   and  Q,  are placed each in a corner of a square of side 2L.  Determine:  (a) The charge Q so that the electric potential at the center of the square is zero. (b) The total electric field at the center of the square. Solution.-

The electric potential is a scalar so its computation is very easy. However, since the electric field is a vector, the electric field of each charge must be expressed as a function of unit vectors   x and  y. In the figure, a wire has been bent so that it has the shape of a semi-circumference of radius R, then the wire has been loaded with an electric charge and, this distributes as a linear density. Determine the electric potential at the origin of coordinates. Solution.-

The linear charge density and the differential of arc are respectively: (8) Two metallic circular disks of radius R are placed isolated, face-to-face, being d the distance between both.  The disks are connected to the terminals of a battery of voltage V, so that they acquire charges of equal magnitude Q but opposite signs.  Determine (a) the capacitance (C = Q / V) for the system (b) Diameter of the disks.

Solution.- A particle with mass m and electric charge Qo is released from rest in a uniform electric field of magnitude Eo. For the moment the particle has undergone a displacement d in the electric field, determine (a) the acceleration of the particle (b) The change in electric potential of the particle. (c) The change in potential energy of the particle. (d) The speed of the particle.

Solution.-

Since the particle is positively charged it will be accelerated in the direction of the electric field and it will displace in that direction, moving from a higher to a lower potential.   Along this displacement the charged particle will increase its kinetic energy, at the cost of its potential energy. After traveling the distance d the kinetic energy of the particle will be higher than its initial one and its potential energy will be lower than the initial. Anatomía del Caos:   Estudio del Caos en modelos matemáticos, basado en Física Computacional y Simulacional. Este libro, escrito por Javier Montenegro Joo, se llama Anatomía del Caos, porque muestra como extraer tomografías longitudinales y transversales de eventos caóticos en el oscilador no-lineal amortiguado y forzado

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 Three point charges whose magnitudes are respectively Qo, 2Qo and 3Qo, are placed each at a vertex of an equilateral triangle of side 2L. Determine: (a) The electric potential at the center of each side of the triangle (b) The potential energy of a particle with charge Q placed at the center of each side.

Solution.-

The sketch below shows the equilateral triangle of side 2L with the electric charges on its vertexes, the mid points on the sides of the triangle are A, B and C, respectively: Three electric charges -Qo, -2Qo and 3Qo, are placed each at a vertex of an equilateral triangle of side 2L. Determine: (a) The electric potential on the sides of the triangle, at half the way between vertexes (b) The potential energy of a particle with charge -Qo placed on each side of the triangle at half the way between vertexes.

Solution.-

The sketch below shows the equilateral triangle of side 2L with the electric charges on its vertexes, the points on the triangle sides at half way between its vertexes are A, B and C, respectively: EconoPhysics: is the new field of knowledge that applies some techniques of Physics to Economy. In this work the author applies the elementary concepts of displacement and velocity to the US-dollar exchange rate and constructs the corresponding Space State and, detects chaotic dynamics in the x-rate.

  The electrical potential at a radial distance r of an R-radius copper sphere, with electric charge, is given by Determine the electric charge of the sphere, (a) Superficial (b) Interior (c) Total

Solution.- In the figure, an electric charge Q is uniformly distributed along a thin rod of length L. Determine the electrical potential at points P1 and P2. Solution.- The sketch shows a b-side cube, two of whose vertices have respectively Q1 and Q2 electric charges. Determine the work to be done to move a Qo charge from vertex 'A' to 'B', following the path shown. Solution.- A compact sphere of radius R has electric charge Qo , determine the electric potential: (a) at a radial  distance r inside the sphere. (b) At the center of the sphere.

Solution.-

This problem is complicated due to the fact that the sphere is not a conductor. In order to compute the potential, the electric fields outside and inside the sphere are needed. The electric field out of the sphere involves all the charge in the sphere. The inner electric field is due to a fraction of the charge in the sphere, this demands using the volumetric charge density. 