AP Physics 1 - Fundamentals of Electric Charge

An electric field line is point from the left towards the right. Where will an electron move when placed in the field?

Towards the left

Towards the right

Upward

Downward

It will not move

Explanation

The electron will move towards the left because electric field lines always point towards the negative charge. The electron is negatively charged and will oppose the negative electric field on the right and move towards the positive end on the left.

Therefore the correct answer is that the electron will move to the left.

Vt physics 10 29 question 7 charges

Consider the given diagram of two electrical charges. Which of the following is true about charges A and B?

Both charges are positive

Both charges are negative

A is positive and B is negative

A is negative and B is positive

Explanation

In this question, we're presented with a diagram in which two electrical charges have field lines pointing away from them. We're then asked to determine a true statement regarding these two charges.

To be able to answer this question correctly, we'll have to recall that for positive electrical charges, the field lines will always point away from the charge. For negative electrical charges, the field lines point inwards toward the charge. Since both charges A and B in the diagram have their field lines pointing away from them, both of them must be positively charged. Also notice that their field lines will not cross one another; instead, they are repelled from one another.

A point charge of charge $q = 45\mu C$ is placed in a uniform electric field of strength $78 \frac{N}{C}$ going to the right. The charge is then moved at an angle of $30^{\text o}$ below the horizontal to the right. What is the net work done on this charge?

$3.04\cdot10^{-3}J$

$-1.56\cdot10^{-3}J$

$1.76\cdot10^{-3}J$

$-6.04\cdot10^{-3}J$

$2.36\cdot10^{-3}J$

Explanation

The work done by an electric field on a charged particle is:

$W=qEcos(\theta)$

Where $\theta$ is the angle between the electric field vector and the displacement vector. Plug in the given values.

$3.04\cdot10^{-3}J$

This answer makes sense since an electric field does work on a positive charge when going in the direction of the field lines.

A student has a neutrally charged glass rod and a neutrally charged silk cloth. When the student rubs the silk cloth on the glass rod, the rod acquires a net positive charge. What is the charge on the silk cloth after the student performs this experiment?

The silk cloth has a charge that is equal in magnitude to the glass rod's charge, but is negative

The silk cloth has no charge

The silk cloth has the same charge as the glass rod

The silk cloth is negatively charged and has a smaller magnitude of charge than the glass rod

The silk cloth is positively charged and has a greater magnitude of charge than the glass rod.

Explanation

Charge is fundamentally conserved. In order to give the glass rod a positive charge, the silk cloth removes electrons from the glass rod. Every electron that leaves the glass rod leaves behind the proton (positively charged particle) that used to balance the negative charge of the electron (negatively charged particle). Every electron that leaves the glass rod goes onto the silk cloth, giving it the same amount of charge as the glass rod, but with a negative sign (since electrons are negatively charged). Only a tiny fraction of the total electrons in the glass rod are removed, even with vigorous charging.

If the test charge is increased by a factor of 4, what happens to the electric potential of that charge?

Remains unchanged

Increases by a factor of

Decreases by a factor of

Explanation

Remember that the formula for the electric potential is given by:

$V=\frac{kQ}{r}$

Where is Coulomb's constant, is the source charge, and is the distance. This formula indicates that electric potential is unaffected by changing the test charge.

By how much will the electric field change if the charges of all source charges are increased by a factor of , and the test charge is increased by a factor of ?

The electric field will be changed by a factor of

The electric field will be changed by a factor of $\frac{1}{2}$

Explanation

The electric field for point charges is given by:

$E=\sum_{1}^{n}\frac{kQ_n}{r^2}$

Where is Coulomb's constant, is the charge of each source charge , is distance of the test charge from the source charge, and is the number of source charges.

In this problem, since all of the source charges are increased by a factor of 2, the electric field will also increase by a factor of 2. The increase in the charge of the test charge is not applied to the strength of the electric field, since the electric field is only dependent on the values and locations of the source charges.

Two charges of and are placed away from each other. Another charge is placed in between these two charges. Where should this third charge be placed in order for it to feel no net electrostatic force?

from the charge

There is not enough information given in this problem to solve the question

from the charge

Explanation

The electrostatic force between two charges is:

$F = k\frac{q_{1}q_{2}}{r^{2}}$

The force that the 3rd charge feels from the charge is:

$F = k\frac{2qq_{3}}{r^{2}_{13}}$

The force that the 3rd charge feels from the charge is:

$F = k\frac{4qq_{3}}{r^{2}_{23}}$

We can also rewrite the distance between the charges as and . In this explanation, we will use as the distance between the and the third unknown charge making the distance between the and the 3rd charge . Finally, we solve for a net force of 0 giving us this relationship between the two opposing forces.

$k\frac{2qq_{3}}{x^{2}} = k\frac{4qq_{3}}{(.5 - x)^{2}}$

Solve for .

$\frac{2}{x^{2}} = \frac{4}{(.5 - x)^{2}}$

A point charge of coulombs experiences a force of in an electric field. What is the magnitude of this electric field?

$150 \frac{N}{C}$

$600 \frac{N}{C}$

$300 \frac{N}{C}$

$100 \frac{N}{C}$

$200 \frac{N}{C}$

Explanation

The formula for the force on a point charge in an electric field is as follows:

is the force on the charge, is the magnitude of the charge and is the electric field. Substituting for our values we obtain:

$E= \frac{300N}{2C}= 150\frac{N}{C}$

Therefore the correct answer is $150 \frac{N}{C}$

Spheres

Two identical conducting spheres are attached to insulated posts and charged so that $Q_{1}=-50nC$ and $Q_{2}=+10nC$ .

The spheres are brought together so they touch, then moved back apart. What is the charge on $Q_{2}$ now?

$Q_{2}=-20nC$

$Q_{2}=-50nC$

$Q_{2}=+20nC$

$Q_{2}=0nC$

Explanation

When the spheres make contact, charges are exchanged. The charges on the spheres will move due to the Coulomb forces from all the particles in each sphere. They will move toward equilibrium.

In this case $Q_{1}$ has a net negative charge while $Q_{2}$ has a net positive charge. will have to migrate to $Q_{2}$ so that $Q_{1}=Q_{2}=-20nC$ .

When the spheres are separated they will have equal charge and the charge on $Q_{2}$ will still be .

Determine the total electric charge of molecules in solution in Coulombs.

Explanation

In solution, the ions will disassociate into and . Since there will be molecules of with a total charge $q_{Na}=1.6*10^{-18}C$ and molecules of with a total charge of $q_{Cl}=-1.6*10^{-18}C$ , the overall charge will be $q_{Total}=q_{Na}+q_{Cl}=0C$

Fundamentals of Electric Charge

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