08 February 2011

Electricity

Problem 1
Two small spheres spaced 35.0cm apart   have equal charge.  How many excess electrons must be present on each sphere if the  magnitude of the force of repulsion between them is ?

Solution
The Coulomb force between two equal charges (spheres) is
We know : , , electrostatic constant: . Then we can find q :
Then we can find the number of excess electrons (since we know the charge of a single electron :
 

Problem 2
A point charge is at the point , , and a second point charge is at the point , .  Find the magnitude and direction of the net electric field at the origin.

Solution
The magnitude of electric field due to charge 1 is
The magnitude of electric field due to charge 2 is
 
To find the net electric field we need to find the vector sum of electric field due to charge 1 and electric field due to charge 2. It is easier to work in term of components. So the x-component of the net electric field is the sum of x-components of electric field due to charge 1 and due to charge 2, and the y-component of the net electric field is the sum of y-components of electric field due to charge 1 and due to charge 2.
 
The x and y components of electric field due to charge 1 are
Both components are positive since the charge is negative.
 
The x and y components of electric field due to charge 2 are
It is negative since the charge is positive.
 
Then we can find the x and y components of the net electric field:
Then we can find the magnitude of the net electric field:
The direction of electric field is determined by the angle between the vector of electric field and axis x:

Problem 3
What must the charge (sign and magnitude) of a particle of mass 5 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 800 N/C?

Solution
We have equilibrium. It means that the net force is 0. Then the gravitational force should be equal to electric force:
From this equation we can find the charge of the partice
The direction of electric force should be upward (since the gravitational force has downward direction). Since the direction of electric field is downward then the electric charge should be negative:
 

Problem 4
What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Solution
The electric force is equal to gravitation force. So we can write:
From this equation we can find the electric field

Problem 5
A particle has a charge of -8.00 nC. Find the magnitude and direction of the electric field due to this particle at a point 0.5 m directly above it.

Solution
The magnitude of electric field due to point charge is given by the expression:
Since the charge is negative then the direction of electric field is downward (toward the charge). 

Problem 6
Two particles having charges of 0.70 nC and 12 nC are separated by a distance of 2 m. At what point along the line connecting the two charges is the net electric field due to the two charges equal to zero?  

Solution
Introduce the distance a from the charge 0.7 nC to the point P, where the net electric field is 0. The net electric field at point P is the sum of two contributions: electric field due to charge 0.7 nC and electric field due to charge 12 nC. They have opposite directions (since both charges are positive). Then the condition that the net electric field is 0 is the following:
where r=2 m is the distance between the charges. From this equation we can find the distance a :
 
Problem 7
A closed surface encloses a net charge of 10 nC . What is the net electric flux through the surface?

Solution
From Gauss's law the flux through the surface is
where  

Problem 8
Each square centimeter of the surface of an infinite plane sheet of paper has excess electrons.  Find the magnitude and direction of the electric field at a point 6.00 cm from the surface of the sheet, if the sheet is large enough to be treated as an infinite plane.

Solution
Electric field due to infinite plane:
Where . The surface charge density (charge density per square meter) is:
Then the magnitude of electric field is
 
The direction of electric field is toward the plane (since the plane is negative charged).


Problem 9.
  A thin disk with a circular hole at its center, called an annulus has inner radius a nd outer radius . The disk has a uniform positive surface charge density . Find the total electric charge on the annulus.

Solution
By definition the total electric charge is equal to the product of the surface charge density and the area of the surface. The area is
Then the total charge is
 




Problem 10.
  A thin disk with a circular hole at its center has inner radius a nd outer radius . The disk has a uniform positive surgace charge density on its surface. The disk lies in the yz plane, with its center at the origin. For an arbitrary point on the x axis (the axis of the disk) find the magnitude of the electric field.

Solution
Disk can be considered as a combination of many rings. Electric field due to a ring of radius R has the following expression:
The linear charge density of a ring is . To find the net electric field we just need to integrate the above expression from a nd :
 

Problem 11.
  A thin disk with a circular hole at its center has inner radius a nd outer radius . The disk has a uniform positive surgace charge density on its surface. The disk lies in the yz plane, with its center at the origin.
A point particle with mass m and negative charge -q is free to move along the x axis (but cannot move off the axis). The particle is originally placed at rest at x=0.01 and released. Find the frequency of oscillations.

Solution
From the previous problem we know the expression for electric field
At small x we have:
Then the force acting on a negative charge (-q) is
This force has a form for which the frequency is . Then the frequency is
 


Problem 12.
  Suppose that of hydrogen is separated into its constituent protons and electrons, and that the protons are all grouped together at one point, and the electrons at another point, 3 meter distant. What is the magnitude of the force between the two groups of charge?

Solution
The first step is to find how many protons (and correspondingly electrons) there are in of hydrogen. To find this number we just need to divide the total mass by the mass of a single proton:
Then the charge of all protons grouped is (where is the proton charge), while the charge of all electrons grouped is . Then the magnitude of the Coulomb force is
 

Problem 13.
  Two small beads having positive charges 25q and q are fixed at the opposite ends of a horizontal insulating rod, extending from the origin (the location of the larger charge) to the point x = d. A third small charged bead is free to slide on the rod. At what position is the third bead in equilibrium?

Solution
Let us assume that the position of the third charge is x (the distance from the larger charge) and the charge of the third charge is Q. Then the equilibrium condition is: the net force is 0. This condition can be written as


From this equation we can find the distance x:

Problem 14.
  Each of the protons in a particle beam has a kinetic energy of . What are the magnitude and direction of the electric field that will stop these protons in a distance of 2 m?

Solution
To stop the proton the direction of electric field should be opposite to the direction of the beam.
To find the magnitude of electric field we need to write down the energy conservation: the initial energy ( K ) of proton should be equal to the work done by an electric field:
Where . From this equation we can find electric field:


Problem 15.
Two identical spheres are each attached to a rope of length 1 m and hung from a common point. Each sphere has mass 1 kg. The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. The spheres have positive charges. The spheres are in equilibrium and each rope makes an angle 30 degrees with the vertical. Find the tension of the ropes.

Solution
The system is in equilibrium. It means that the net force is 0. It means that the sum of the tension force and the gravitational force should have horizontal direction. Then we can write:
 

From this equation we can find the tension:
 

Problem 16.
A solid metal sphere with radius 0.75 m carries a net charge of 0.13 nC. Find the magnitude of the electric field at the following locations:
a) at a point 0.15 m outside the surface of the sphere.
b) at a point outside the sphere, 0.1 m below the surface.

Solution
The specific property of conductor (metal) is that the electric field inside conductor (in equilibrium) is 0. From this property it is possible to obtain that the charge is distributed over the surface of conductor (not in the bulk).
In the present problem the conductor has the shape of sphere. It means that all the charge 0.13 nC will be on the surface of the sphere.
Then the electric field outside of the conductor is the same as the electric field of the point charge at the center of the sphere.
Then electric field inside of the conductor is 0.

(a) a point 0.15 m outside the surface of the sphere is at distance 0.15+0.75 = 0.9 m from the center of the sphere. Then electric field at this point is

(b) Electric field inside of the sphere (inside of the conductor) is 0. 

Problem 17.
How many excess electrons must be added to an isolated spherical conductor 27 cm in diameter to produce an electric field of 1450 N/C just outside the surface?

Solution
The electric field just outside of the spherical conductor is
where is the radius of the spherical conductor.
We know the electric field. Then we can find the excess charge:

To find the number of excess electrons we just need to divide the excess charge by the charge of a single electron:

Problem 18.
To study the structure of the lead nucleus, electrons are fixed at a lead target. Some of the electrons actually enter the nuclei of the target, and the deflection of these electrons is measured. The deflection is caused by the charge of the nucleus, which is distributed approximately uniformly over the spherical volume of the nucleus. A lead nucleus has a charge of +82e ( ) and a radius of . Find the acceleration of an electron at the following distances from the center of a nucleus.
(a) R
(b) 2R
(c) R/2
(d) 0 (at the center)

Solution
In the present problem we have a uniformly charged sphere. The electric field of such sphere has the following expression (Q is the charge of the sphere):

Then the electric force on an electron is

The acceleration is the ratio of the force and the mass of the electron:
where .

Then
(a) at r=R:

(a) at r=2R:

(a) at r=R/2:

(a) at r=0:

Problem 19.
An electrostatic field is given by  
Determine the work done in moving the charge
(a) from (0,0,0) to (4,0,0) m
(b) from (4,0,0) to (4,2,0) m
(c) from (4,2,0) to (0,0,0) m along a straight path.

Solution
If the charge is moving from point A to point B then the work done by electric field is given by the expression:
In this expression the integral is calculated along any trajectory connecting points A and B.
 
(a) We calculate the integral along the straight line connecting initial and the final points:

(b) We calculate the integral along the straight line connecting initial and the final points:

(c) We calculate the integral along the straight line connecting initial and the final points:


Problem 20.
A cylinder with radius r= 0.75 m and length l=0.6 m that has an infinite line of positive charge running along its axis.  The charge per unit length on the line is .
(a) What is the electric flux through the cylinder due to this line of charge?
(b) What is the flux through the cylinder if its radius increased to r= 0.32 m?
(c) What is the flux through the cylinder if its length increased to l= 0.8 m?

Solution
From Gauss's law the flux through the surface depends only on the charge inside of this surface.
where
 
(a) The charge inside of the cylinder depends only on the length of the cylinder. It is
Then

(b) The charge inside of the cylinder depends only on the length of the cylinder, so in this case the charge is the same as in case (a). So the flux will be the same:

(c) Now the length is 0.8 m. Then
Then

Problem 21.
Three small spheres are enclosed in surface carrying charges q1 = 4 nC, q2 = -10.00 nC, and q3 = 5 nC.
 Surface (s)                        What it encloses
s1                                        q1
s2                                        q2
s3                                        q1 and q2
s4                                        q1 and q3
s5                                        q1, q2 and q3


Find the net electric flux through each of the following closed surfaces.
(a) s1
(b) s2    
(c) s3        
(d) s4
(e) s5

Solution
From Gauss's law the flux through any surface depends only on the charge inside of the surface
Then
(a)
where

(b)

(c)

(d)

(e)  
Problem 22.
Two plastic spheres, each carrying charge uniformly distributed throughout its interior, are initially placed in contact and then released. The first sphere has a diameter of 60 cm, a mass of  50 g, and contains   of charge. The second sphere has a diameter of 40 cm, a mass of  150 g and contains    of charge. Assume that no other forces are acting on them.
(a) Find the maximum speed achieved by the first sphere.
(b) Find the maximum acceleration achieved by the first sphere.

Solution
(a) The initial distance between the spheres (between the centers of the spheres) is 30 cm+20cm=50cm=0.5 m. The initial velocities of both spheres are zero. Then we release the spheres and due to repulsion between them they will move away from each other. Finally when the distance between them is very large (infinite) the velocities will be maximal.
To find these velocities we need to write down the energy and momentum conservation laws. If is the final velocity of the first sphere, is the final velocity of the second sphere then we have:


From the second equation we can find the speed of the second sphere, then substitute it in the first equation and find the speed of the first sphere:


Then


 
(b) The maximum acceleration is achieved when the distance between the spheres is the smallest. Then


Problem 23.
A particle with a charge of +8 nC is in a uniform electric field E directed to the left. It is released from rest and moves to the left. After it has moved 3 cm, its kinetic energy is found to be
(a) What work was done by the electric force?
(b) What is the potential of the starting point with respect to the endpoint?
(c) What is the magnitude of E?

Solution
(a) The work done by electric force is exactly equal to the kinetic energy of the particle. So work is

(b) At the same time we can write the work done by electric force as the change of potential energy (or potential multiplied by the charge of the particle):

Then


(c) The potential difference is equal to E*d, where d is the traveled distance. Then



Problem 24.
A total electric charge of 8 nC is distributed uniformly over the surface of a metal sphere with a radius of 27 cm. If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere.
(a) 52 cm
(b) 20 cm
(c) 9 cm
 
Solution
The potential has the following value:

Then for R=27 cm we have:

(a) 52 cm>R then

(b) 20 cm

(c) 9 cm
 
Problem 25.
A potential difference of 10 kV is established between parallel plates in air.
(a) If the air becomes electrically conducting when the electric field exceeds , what is the minimum separation of the plates?
(b) When the separation has the minimum value calculated in part (a), what is the surface charge density on each plate?


Solution
(a) The potential difference is related to the electric field by the following equation:
From this equation we can find the minimum distance:


(b) The electric field between two parallel plates has the following form
Then


Problem 26.
An alpha particle with kinetic energy 10 MeV makes a head-on collision with a lead nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)


Solution
In this problem we need to use energy conservation. The initial energy (kinetic energy of alpha particle) should be equal to the final energy, which is the potential energy of interaction of alpha particle and a lead nucleus. Then

We know the initial energy:

We know the charge of the particles:


Then we can find the shortest distance:
 
Problem 27.
In a certain region of space, the electric potential is , where A, B, and C are positive constants.
(a)  Calculate the x-, y-, and z-components of the electric field.
(b) At which points is the electric field equal to zero?


Solution
(a) By definition the x-,y- and z-components of electric field are


(b) The electric field is zero if all its components are zero. Therefore:

From these equations we can find:


Problem 28.
A point charge is held stationary at the origin. A second point charge is moves from the point x = 0.20 m, y = 0 to the point x = 0.3 m, y = 0.35 m. How much work is done by the electric force on ?


Solution
The work done by electric force on the second charge is equal to (minus) change of the electric potential energy of the charge in the field of charge .

The initial distance between the charges is

The final distance is

Then

Problem 29.
A parallel-plate air capacitor has a capacitance of 800 pF. The charge on each plate is 6 µC. What is the potential difference between the plates?


Solution
In this problem we just need to use the relation between the capacitance of the capacitor, its charge and the potential difference between the plates:

From this equation we can find:

Problem 30.
A parallel-plate air capacitor of capacitance of 100 pF has a charge of magnitude 0.1 µC on each plate. The plates are 0.5 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?


Solution
(a) To find the potential difference we need to use the equation:
From this equation we can find:

(b) To find the area of the plates we need to use the expression for the capacitance:
We know d=0.5 mm=0.0005 m, then we can find the area A:

(c) The potential difference is related to the electric field through the following relation:
Then

(d) The surface charge density can be found from the equation:

Problem 31.
In a Millikan oil drop experiment a student sprayed oil droplets with a density of between two horizontal parallel plates that were 6.0 cm apart. The student adjusted the potential difference between the plates to 5000 V so that one of the drops became stationary. The diameter of this drop was measured to be . What was the magnitude of the charge on the drop.


Solution
In the stationary state the gravitational force should be equal to electric force. The gravitation force is
where d is the diameter of the droplet.

This force should be equal to electric force, which is
 
Then
 
Problem 32.
You want to create an electric field at location . Where would you place a proton to produce this filed at the origin? Instead of a proton, where would you place an electron to produce this filed at the origin?


Solution

We have only y-component of electric field. It means that the charge should be placed along axis y. Since the y-component of electric field is positive and the proton is positively charge particle then the coordinate of the proton should be negative. The magnitude of electric field is
Where y is the coordinate of the proton. Then from this equation we can find y and we remember that y is negative:


Since the charge of an electron is negative then the coordinate of electron should be positive (to produce the positive y-coordinate of electric field. The magnitude of the electron charge is the same as the magnitude of the proton charge. Then the position of electron is
 
 

Problem 33.
A cube of side 1 cm is placed within an electric field of  100 N/C. What is the flux through the (a) top face (b) bottom face (c) right face (d) left face of the cube. What is the net flux through the cube. The electric field is perpendicular to the left face of the cube.


Solution

To characterize the orientation of the faces of the cube we introduce the normal to the surface (faces) of the cube as shown in the figure below. Then the flux through any surface can be calculated as
where is the magnitude of the electric field, is the area of the surface, and is the angle between the direction of the electric field and the direction of the normal to the surface.



(a) Top face: The normal is perpendicular to the electric field. Then and . Then the flux is 0.
(b) Bottom face: The normal is perpendicular to the electric field. Then and . Then the flux is 0.
(c) Right face: The normal to the right face has the same direction as the direction of the electric field. Then and . The area of the face of the cube is (we need to use the correct units for the side of the cube: 1 cm = 0.01 m):
Then the flux is
(d) Left face: The direction of the normal to the left face is opposite to the direction of the electric field. Then and . The area of the face of the cube is (we need to use the correct units for the side of the cube: 1 cm = 0.01 m):
Then the flux is
 
The net flux through the cube is proportional to the net charge inside the cube (Gauss's law). Since the charge inside the cube is zero then the net flux is zero. 

Problem 34.
A particle of charge Q and mass m is accelerated from rest through a potential difference V, attaining a kinetic energy K. What is the kinetic energy of a particle of charge 2Q and mass m/2 that is accelerated from rest through the same potential difference?

Solution
We need to use the energy conservation law: the work done by electric force is equal to increase of the kinetic energy of the particle. The work done by electric force is equal to the product of the charge of the particle and the potential difference: . Then the kinetic energy is
.......................................................................(1)
(the initial kinetic energy is zero).
 
For the particle with charge 2Q and mass m/2 we have the same expression but know with the charge of the particle 2Q:
Taking into account expression (1) we obtain:
 
Interesting, the result does not depend on the mass of the particle. 

Problem 35.
Initially, sphere A has a charge of -50e and a sphere B has a charge of +20e. The spheres are made of conducting material and are identical in size. If the spheres then touch, what is the resulting charge on sphere A?


Solution
In this problem we need to use two facts:
(1) the conservation of the net charge of the system of two spheres and
(2) the fact that the spheres are conducting, which means that the potentials of the spheres should be the same (at the moment when they touch each other). Since the spheres are identical in size then the fact that the potential of the spheres are the same means that the charges of the sphere (when the spheres are in contact) should be the same.

Therefore the final charge of sphere A is equal to the final charge of sphere B:

 

The conservation of the charge of the system of two spheres (this is an isolated system) means that the net initial charge of two spheres is equal to the net final charge:
 

We know the initial charges of the spheres, then



Problem 36.
A cube of edge a carries a point charge q at each corner. Find the resultant electric force on any one of the charges.


Solution
We labels four identical charges as: Charge 1, Charge 2, Charge 3, and Charge 4 (as shown in the figure). We calculate the net electric force on charge 1.

There are three electric forces acting on charge 1:
1. due to charge 2. The magnitude of the force is
The distance between the charges 1 and 2 is . Then the force is

2. due to charge3. The magnitude of the force is
The distance between the charges 1 and 3 is . Then the force is

3. due to charge4. The magnitude of the force is
The distance between the charges 1 and 4 is . Then the force is


All these forces are vectors. The directions of the forces are shown in the figure. To find the net force – the vector sum of all three forces – we need to introduce the coordinate system: x and y.

Then the x and y components of the net force are

From the known magnitudes and the directions of the forces we can find the corresponding components:

Then the magnitude of the net force is



AC current

Problem 1.

A capacitance of is connected to an alternating emf of frequency 100 cycles/second. What is the capacitive reactance?

Solution: 
The capacitive reactance is determined by the following expression:
where – frequency, and – capacitance. In the above expression we need to use the correct units for the capacitance: F (farad). Then .
Now we can find the capacitive reactance: