### All High School Physics Resources

## Example Questions

### Example Question #6 : Electric Charge

A neutral atom always has

**Possible Answers:**

The same number of protons and electrons

Possible Answers

More protons than electrons

The same number of neutrons and protons

More neutrons than protons

**Correct answer:**

The same number of protons and electrons

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.

### Example Question #7 : Electric Charge

As an object acquires a positive charge, its mass usually

**Possible Answers:**

Stays the same

Becomes negative

Increases

Decreases

**Correct answer:**

Decreases

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).

### Example Question #8 : Electric Charge

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

**Possible Answers:**

**Correct answer:**

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

### Example Question #9 : Electric Charge

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

**Possible Answers:**

The field points away from the positive charge, and the force on the negative charge is in the opposite direction as the field

The field points away from the positive charge, and the force on the negative charge is in the same direction as the field

The field points toward the positive charge and the force on the negative charge is in the same direction as the field

The field points toward the positive charge and the force on the negative charge is in the opposite direction as the field

**Correct answer:**

The field points away from the positive charge, and the force on the negative charge is in the opposite direction as the field

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.

### Example Question #10 : Electric Charge

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?

**Possible Answers:**

**Correct answer:**

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

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

### Example Question #931 : High School Physics

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.

**Possible Answers:**

**Correct 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

Rearranging this we get

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

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

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

### Example Question #932 : High School Physics

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.

**Possible Answers:**

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

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

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

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

**Correct answer:**

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

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

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 .

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.

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

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

We can set these two equations equal to each other.

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

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

Cross multiply on both sides

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

So the two options for the value would be

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

We can plug in our value for the .

The and can be cancelled out

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

### Example Question #933 : High School Physics

Particles of charges , , and are placed in a line. The center charge is and is away from each of the others. Calculate the net force on the center charge from the other two.

**Possible Answers:**

**Correct answer:**

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

The first charge on the middle charge

The last charge on the middle charge

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

### Example Question #934 : High School Physics

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?

**Possible Answers:**

**Correct 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.

### Example Question #935 : High School Physics

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

**Possible Answers:**

reduced by a factor of 4

doubled

reduced by a factor of 2–√

reduced by a factor of 3

quadrupled

**Correct answer:**

reduced by a factor of 4

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.