Electricity - High School Physics

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Question

How many electrons make up a charge of $-48\mu C$ ?

Answer

The charge of a single electron is $-1.6x10^{-19} C$

We can then convert the amount of charge to determine the number of electrons.

$-48\mu C * \frac{1C}{1x10^{6}\mu C} * \frac{1 electron}{-1.6x10^{-19}C} = 3x10^{14} electrons$

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Question

A neutral atom always has

Answer

Protons are positively charged and electrons are negatively charged. In order for an object to be considered neutral, it must have the same number of both positive and negative particles. Therefore it must have the same number of both protons and electrons.

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Question

Materials in which the electrons are bound very loosely to the nuclei and can move about freely within the material are referred to as

Answer

Conductors allow the electrons to flow freely along it. That is why metal; is considered a good conductor. It allows the electrons to flow through it which is why it is used in wire in an an electric circuit.

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Question

A glass rod is rubbed with a piece of silk. During the process the glass rod acquires a positive charge. The silk

Answer

Since the glass rod acquires a positive charge, this means that it is deficient in electrons. Since the rod was rubbed by the piece of silk, the silk is what now collects the electrons. The silk now has an excess of electrons which means that the silk is now negatively charged.

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Question

A charged rod carrying a negative charge is brought near two spheres that are in contact with each other but insulated from the ground. If the two spheres are then separated, what kind of charge will be on the spheres?

Answer

When the negatively charged rod is brought near one of the two spheres, the presents of the negative charge will induce a flow of charge in the spheres such that regions farthest away from the charged rod will become most negative and regions near the rod will become most positive. This is called charge by induction.

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Question

By what method will a positively charged rod produce a negative charge on a conducting sphere that is placed on an insulating surface?

Answer

Charge by induction happens when a charged object is brought in the vicinity of a neutral object. The presents of the charged object will cause the free charges in the neutral object to shift such that the neutral object becomes polarized. When the charged object is positive, this will induce a negative charge on a neutral object.

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Question

As an object acquires a positive charge, its mass usually

Answer

When an object acquires a positive charge, it is losing electrons to its surroundings. If it is losing electrons, it is losing mass (although ever so slightly).

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Question

What is the net charge of a bar of gold? Gold has electrons per atom and an atomic mass of .

Answer

Since a gold atom has balanced protons and electrons, the net charge on the atom is .

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Question

A negative point charge is in an electric field created by a positive point charge. Which of the following is true?

Answer

By convention, electric fields point away from positive charges and toward negative charges. Since the field is created by a positive point charge, the electric field will point away from the positive charge.

Negative and positive charges attract one another. Therefore the force on the negative charge is toward the positive charge. This is in the opposite direction as the field.

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Question

When an object such as a plastic comb is charged by rabbit it with a cloth, the net charge is typically a few microcoulombs. If that charge is , by what amount does the mass of a comb change during charging?

Answer

First, we need to determine how many additional electrons are on the comb due to the of charge.

$3C \frac{1 electron}{1.610^{-19}} = 1.87510^{19} electrons$

Now that we know the number of electrons, we can determine the mass of these additional electrons.

$1.87510^{19} electron \frac{9.1110^{-31}kg}{1 electron} = 1.7*10^{-11}kg$

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Question

At each corner of square of side there are point charges of magnitude Q, 2Q, 3Q and 4Q. Determine the magnitude and direction of the force on the charge 3Q.

Answer

We will need to use Coulomb’s Law to analyze the force on the 3Q charge from all the other forces. We will then summarize the net force in the and direction to determine the force on the 3Q charge.

In the x-direction

Force from 4Q on 3Q

$F = \frac{k4Q3Q}{l^2}$

Force from Q on 3Q in the x direction

$F =\frac{kQ3Q}{(\sqrt{2}l)^2}$

$F = \frac{k3Q^2}{2l^2}$

Add these together in the x-direction

$F = \frac{k12Q^2}{l^2} + \frac{k3Q^2}{2l^2}$

$F = \frac{27kQ^2}{2l^2}$

In the y-direction

From from 2Q on 3Q

$F = \frac{k2Q3Q}{l^2}$

$F = \frac{k6Q^2}{l^2}$

Force from Q on 3Q in the y direction

$F =\frac{kQ3Q}{(\sqrt{2}l)^2}$

$F = \frac{k3Q^2}{2l^2}$

Add these together in the y-direction

$F = \frac{k6Q^2}{l^2} + \frac{k3Q^2}{2l^2}$

$F = \frac{15kQ^2}{2l^2}$

Now we can find the resultant of these sides using the Pythagorean theorem.

$\frac{27kQ^2}{2l^2}^2 + \frac{15kQ^2}{2l^2}^2 = c^2$

$\frac{729k^2Q^4} {4l^4} + \frac{225k^2Q^4}{4l^2} = c^2$

$\frac{954k^2Q^4}{4l^4} = c^2$

$\sqrt{\frac{954k^2Q^4}{4l^4}} = \sqrt{c^2}$

$\frac{30.9kQ^2}{2l^2} = c$

$15.5\frac{kQ^2}{l^2} = c$

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Question

A large electroscope is made with “leaves” that are long wires with negligible mass with tiny spheres at the end. When charged, nearly all the charge resides in the sphere. If the wires each make a angle with the vertical, what total charge must have been applied to the electroscopes.

Answer

Each of these charged spheres exert a force on each that is equal and opposite on one another in the x-direction. The tension force on each of these charged spheres is equal to the force from the other sphere in the x-direction and the force of gravity in the y-direction. We will be able to use this information to determine the charge on the electroscope.

First we must analyze the forces on the charges and determine the force on the charges. We know that

$tan\theta = {F_{sphere}}{F_{gravity}}$

Rearranging this we get

$F_{gravity}tan\theta = F_{sphere}$

$(0.020kg)(9.8m/s^2)tan(25) = F_{sphere}$

$F_{sphere} = 0.091N$

Next we can use Coulomb’s law to determine the magnitude of each charge.

$F_{sphere} = \frac{kq_1q_2}{d^2}$

In order to figure out the distance between the two spheres we can trigonometry with the length of the string.

$sin\theta = \frac{opp}{hyp}$

Therefore the distance between the two charges is double this value.

$0.091N = \frac{(9x10^9)(\frac{Q}{2})(\frac{Q}{2})}{(0.68m)^2}$

$0.091N = \frac{(2.25x10^)Q^2}{0.4624}$

$Q = 4.32x10^{-6}C$

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Question

Two charges -Q and -3Q are a distance l apart. These two charges are free to move but do not because there is a third(fixed) charge nearby. What must the magnitude of the third charge and its placement be in order for the first two to be in equilibrium.

Answer

First let us determine the force of the two charges on each other.

$F = \frac{kq_1q_2}{d^2}$

$F = \frac{k(-Q)(-3Q)}{l^2}$

$F = \frac{k3Q^2}{l^2}$

The easiest way to analyze this is to assume all of these are in a straight line and we have a charged placed somewhere along the same x-axis between the charges that keeps them all from moving. Let us assume the smaller charge is at the origin and the distance to the fixed charge is some distance .

$F = \frac{k(-Q)(xQ)}{r^2}$

Let us also assume that the larger charge is a distance l away from the origin (away from the smaller charge) and therefore a distanc l-r away from the fixed charge.

$F = \frac{k(-3Q)(xQ)}{(l-r)^2}$

To be in equilibrium we know that the net force on the smaller charge must equal .

$\frac{k3Q^2}{l^2} + \frac{k(-Q)(xQ)}{(r)^2} = 0$

We also know that the net force on the larger charge must equal

$\frac{k3Q^2}{l^2} + \frac{k(-3Q)(xQ)}{(l-r)^2} = 0$

We can set these two equations equal to each other.

$\frac{k3Q^2}{l^2} + \frac{k(-Q)(xQ)}{(r)^2} = \frac{k3Q^2}{l^2} + \frac{k(-3Q)(xQ)}{(l-r)^2}$

The force between the charges can be cancelled since it is on both sides.

$\frac{k(-Q)(xQ)}{(r)^2} = \frac{k(-3Q)(xQ)}{(l-r)^2}$

The k and Q values can also be cancelled out on both sides.

$\frac{-1}{(r)^2} = \frac{-3(l-r)^2} {[?]}$

Cross multiply on both sides

This is a quadratic that needs to be solved with a quadratic formula.

$r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$r =\frac{-(-2l)\pm\sqrt{(2l)^2-4(-2)(l^2)}}{2(-2)}$

$r = \frac{2l\pm\sqrt{4l^2+8l^2}}{-4}$

$r = \frac{2l\pm\sqrt{12l^2}}{-4}$

$r = \frac{2l\pm2\sqrt{3}l}{-4}$

$r = \frac{l\pm\sqrt{3}l}{-2}$

$r = \frac{-l\pm\sqrt{3}l}{2}$

So the two options for the value would be

$r = \frac{-2.73l}{2} = -1.37l$

$r = \frac{0.73l}{2} = 0.366l$

Of these two, the one that makes the most sense based on our assumptions is the as this would be in the middle of two charges.

We can now go back to our first equation with the smaller charges

$\frac{k3Q^2}{l^2} + \frac{k(-Q)(xQ)}{(r)^2} = 0$

We can plug in our value for the .

$\frac{k3Q^2}{l^2} +\frac{k(-Q)(xQ)}{(0.366l)^2} = 0$

The and can be cancelled out

$3 + \frac{x}{(0.366)^2} = 0$

$3 +\frac{x}{.134} = 0$

$-3 = \frac{x}{.134}$

The charge must be placed away from the smaller charge with a magnitude of .

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Question

What is the magnitude of force a $+25\mu C$ charge exerts on a charge away?

Answer

We can calculate this using Coulomb’s Law

$F = \frac{kq_1q_2}{d^2}$

$F = \frac{(9x10^9)(25x10^{-6})(25x10^{-3})}{(0.16m)^2}$

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Question

Particles of charges $+70\mu C$ , $+45\mu C$ , and $-100\mu C$ are placed in a line. The center charge is $+45\mu C$ and is away from each of the others. Calculate the net force on the center charge from the other two.

Answer

We can calculate the force from each of the two charges on the center charge using Coulomb’s Law

$F = \frac{kq_1q_2}{d^2}$

The first charge on the middle charge

$F = \frac{(9x10^9)(70x10^{-6})(45x10^{-6})}{(0.4m)^2}$

The last charge on the middle charge

$F = \frac{(9x10^9)(45x10^{-6})(-100x10^{-6})}{(0.4m)^2}$

To find the total force we need to add both of these forces together.

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Question

A point charge of +Q is placed at the center of a square. When a second point charge of -Q is placed at one of the square’s corners it is observed that an electrostatic force of 4N acts on the positive charge at the square’s center. Now, identical charges of -Q are placed at the other three corners of the square. What is the magnitude of the net electrostatic force acting on the positive charge at the center of the square?

Answer

Since the charges are opposite, the center charge +Q is going to be attracted to each of the charges in the corner of the square. Since each charge is equidistant from the center charge, they will each exert 4N of force on the charge in the center. However, since each corner charge is pulling the center charge equally and oppositely, the net force on the system is equal to 0 and the charge will not move. All of the forces cancel out with one another.

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Question

When the magnitude of two interacting charges is increased by a factor of 2, the electrical forces between these charges is __________.

Answer

In Coulomb's law, an increase in both interacting charges will cause an increase in the magnitude of the electrical force between them. Specifically if the magnitude of both interacting charges is doubled, this will quadruple the electrical force.

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Question

When the distance of two interacting charges is increased by a factor of 2, the electrical forces between these charges is __________.

Answer

In Coulomb’s Law, an increase in the distance will cause a decrease in the magnitude of the electrical force between them. Since this is an example of an inverse square law, doubling the distance will reduce the force by a factor of 4.

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Question

What is the main difference between electrical and gravitational forces?

Answer

Electric forces can be attractive or repulsive because charges may be positive or negative. In the case for gravitational forces, there are only attractive forces because mass is always positive.

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Question

Calculate the magnitude of the electric field at a point that is located $\small 15cm$ directly north of a point charge, $\small Q = -15.0 * 10^{-6}C$ .

$k=9*10^9\frac{m^2N}{C^2}$

Answer

The magnitude of an electric field due to a single point charge is given by the formula:

$E=k\frac{Q}{r^2}$

The variable $\small k$ refers to Coulomb's constant. We are given the charge and the distance between points. To use the equation, first convert the distance to meters.

$15cm\frac{1m}{100cm}=0.15m$

We can now use the given constant, charge, and distance in the equation to solve for the magnetic field strength.

$E=(910^9\frac{m^2N}{C^2})(\frac{-15.010^{-6}C}{(0.15)^2})$

$E=(919^9\frac{m^2N}{C^2})(-6.6710^{-4}\frac{C}{m^2})$

$E=-6.010^6\frac{N}{C}$

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