Electrostatics and Electrical Fields - MCAT Physical

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Question

Two charges of $Q$ coulombs are a distance $d$ apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Answer

Given Coulomb's Law electrostatic forces: $F= \frac{kQq}{d^{2}}$

We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

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Question

$e = 1.60 \times 10^{-19}$

$k = 8.99 \times 10^{9}$

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.

Answer

To find the answer we must first solve for q, using Coulomb's Law. Then after finding q = 1.85 x 10–6 C, we must divide by 1.6 x 10–19 C to get the amount of electrons required to make such a charge.

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Question

Two charges Q1 and Q2 are held stationary 1 m away from each other. The magnitude of charge of Q2 is twice that of Q1. A positive point charge A is placed directly in the middle of the two charges (Q1 and Q2). What is the ratio of the net force on charge A when Q1 and Q2 are both negative over the net force on charge A when Q1 is negative and Q2 is positive?

Answer

For this problem we must understand Coulomb's Law. The force on charge A will be dircetly affected by the charge of Q1 and Q2. Q1 and Q2 are both negative, they will both attract charge A, and the net force will be reduced as they pull in opposite directions. Switching Q2 to a positive charge will result in a repulsive force on charge A which will be in the same direction as the attractive force between Q1 and A.

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Question

Two point charges, Q1 and Q2, are placed on the x-axis. Q1 is placed at x = 0m and Q2 at x = 1m. If a positive charge, Q3, is placed on the x-axis at x = 0.25m, it experiences a net electric force of 0. Which of the following could be determined based on this observation?

Answer

In order for there to be 0 net force on Q3, the forces from Q1 and Q2 must be in opposite directions. So Q1 and Q2 must either be both attracting Q3 or both repelling Q3 (that is, Q1 and Q2 are either both negative or both positive).

Also, since these forces must have equal magnitude in order to cancel, such that $\small F_{13}=F_{23}$ , where $\small \small F_{13}$ is the magnitude of force between Q1 and Q3, and similarly for $\small F_{23}$ . Recalling Coulomb’s law for electric force, $\small F_{13}=\frac{kQ_{1}Q_{3}}{r_{13}^{2}}$ and $\small \small F_{23}=\frac{kQ_{2}Q_{3}}{r_{23}^{2}}$ (where Q1, Q2, and Q3 are the magnitudes of the respective charges). Since $\small r_{23}>r_{13}$ , it must also be true that $\small kQ_{2}Q_{3}>kQ_{1}Q_{3}$ for the two fractions to have the same value. So, Q2 must have a greater magnitude than Q1.

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Question

Electronegativity is an important concept in physical chemistry, and often used to help quantify the dipole moment of polar compounds. Polar compounds are different from those compounds that are purely nonpolar or purely ionic. An example can be seen by contrasting sodium chloride, NaCl, with an organic molecule, R-C-OH. The former is purely ionic, and the latter is polar covalent.

When comparing more than one polar covalent molecule, we use the dipole moment value to help us determine relative strength of polarity. Dipole moment, however, is dependent on the electronegativity of the atoms making up the bond. Electronegativity is a property inherent to the atom in question, whereas dipole moment is a property of the bond between them.

For example, oxygen has an electronegativity of 3.44, and hydrogen of 2.20. In other words, oxygen more strongly attracts electrons when in a bond with hydrogen. This leads to the O-H bond having a dipole moment.

When all the dipole moments of polar bonds in a molecule are summed, the molecular dipole moment results, as per the following equation.

Dipole moment = charge * separation distance

Electronegativity is based on the principle that the closer an electron is to the atomic nucleus, where postive charge is concentrated, the more attractive force the positive charge can exert on the electron.

As a result, as atomic radius decreases by one half, what happens to the force pulling the electron toward that atom?

Answer

This is a function of the equation $F = k\frac{q_1q_2}{d^2}$ .

In other words, as the distance between two charges doubles, the force between them goes increases, as the denominator decreases from to $(\frac{1}{2}d)^2$ .

Here, we are decreasing the distance by 1/2, decreasing the denominator by 1/4, so our force goes up by 4 times.

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Question

Batteries and AC current are often used to charge a capacitor. A common example of capacitor use is in computer hard drives, where capacitors are charged in a specific pattern to code information. A simplified circuit with capacitors can be seen below. The capacitance of C1 is 0.5 μF and the capacitances of C2 and C3 are 1 μF each. A 10 V battery with an internal resistance of 1 Ω supplies the circuit.

Using the plate diagram above, what direction do the electric field lines point?

Answer

As with a point charge, remember that electric field lines point from areas of high potential to areas of low potential. Areas of high potential have positive (+) charge and areas of low potential have negative (-) charge, thus the electric field lines point to the right.

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Question

Batteries and AC current are often used to charge a capacitor. A common example of capacitor use is in computer hard drives, where capacitors are charged in a specific pattern to code information. A simplified circuit with capacitors can be seen below. The capacitance of C1 is 0.5 μF and the capacitances of C2 and C3 are 1 μF each. A 10 V battery with an internal resistance of 1 Ω supplies the circuit.

Using the plate diagram above, in what direction do the electric force lines point?

Answer

As with a point charge, remember that electric force lines point away from positive charge and towards negative charge. As can be seen in the image, electric force lines point to the right. As an aside, electric field lines point in the same direction as electric force lines, so if you know the direction of one, you also know the direction of the other.

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Question

Which of the following equations best relates capacitance, charge, and distance between parallel plates as they pertain to the electric field, ?

Answer

This question forces us to combine two equations.

$C = \frac{Q}{V}$

Using substitution, we can solve for the electric field with the variables given in the question stem.

$C = \frac{Q}{Ed}$

$E=\frac{Q}{Cd}$

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Question

Consider a spherical shell of radius has a charge .

What is the electric field away from the center of the shell?

Answer

The electric field inside a charged spherical shell is always zero.

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Question

Consider a spherical shell of radius has a charge .

What is the electric field at a distance from the surface of the shell?

Answer

According to the shell theorem, a charged spherical shell can be treated as a point charge at the center of the spherical shell. The equation for the electric field of a point charge is:

$E=\frac{kQ}{r^2}$

In this case, since we are at a distance of from the surface of the shell and therefore away from the shell's center. Plug into the equation for electric field.

$E=\frac{kQ}{(3R)^2} = \frac{kQ}{9R^2}$

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Question

For the following:

$e = 1.60 \times 10^{-19}$

$k = 8.99 \times 10^{9}$

If a charge of magnitude +4e is being held in place 3 nm from a charge of -5e which is also being held in place. What is the potential energy of the system?

Answer

$U= \frac{kQq}{r}$

Given the equation and plugging in the values of e and k, we get that

$U= 1.53\times 10^{-18}$

It is important to keep in mind that the charge $e$ is given in the question and must be incorporated into the formula by multiplying each charge by that value.

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Question

If the electrical potential energy between two equal charges quadruples, describe the change in the distance between the particles.

Answer

Electrical potential energy is given by the equation $U=\frac{kq_1q_2}{r}$ .

Electrical potential energy is inversely proportional to the distance between the two charges. If the energy is quadrupled, then (the distance between the two equal charges) must have decreased proportionally.

$U=\frac{kq_1q_2}{r}$

$4U=\frac{kq_1q_2}{x(r)}$

$4U=\frac{kq_1q_2}{(\frac{1}{4})r}=4\frac{kq_1q_2}{r}$

For the energy to be quadrupled, the radius must be quartered.

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Question

Which of the following is not true regarding electric potential?

Answer

A positive test charge will naturally move from high potential to low potential. If it is moved in the opposite direction, then the electric field will do work against its motion (negative work). This be seen from the equation for electric field work:

$W=-q\Delta V$

$\small W$ is the work done by the electric field, $\small q$ is the charge, and $\small \Delta V$ is the potential difference. If $\small \Delta V$ is positive (the final potential is higher than the initial potential) and $\small q$ is also positive, then work done by the field is negative.

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Question

How much work is required to bring together the three given charges from infinity to the corners of an equilateral triangle of side length 1cm?

$q_1 = 1\mu C,\ q_2=2\mu C,\ q_3=3\mu C$

Answer

Relevant equations:

$U = \frac{kq_1q_2}{r}$

Step 1: Since the work done to assemble the charges equals their potential energy in this arrangment, find the potential energy between each pair of charges. Work is equal to change in potential energy; since the charges start at infinite distance, initially potential energy is equal to zero.

Charges 1 and 2

$U_{12} = \frac{kq_1q_2}{r_{12}}= \frac{(910^9)(110^{-6})(210^{-6})}{0.01}=1.8 J$

Charges 1 and 3*

$U_{13}=\frac{kq_1q_3}{r_{13}}=\frac{(910^9)(110^{-6})(310^{-6})}{0.01}=2.7J$

Charges 2 and 3*

$U_{23} = \frac{kq_2q_3}{r_{23}}=\frac{(910^9)(210^{-6})(3*10^{-6})}{0.01}=5.4J$

Step 2: Add together all these potential energies to find the total energy of the arrangement.

$U = U_{12}+U_{13}+U_{23}= 1.8+2.7+5.4 = 9.9J$

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Question

A standard AC outlet is capable of delivering $120V_{rms}$ of current. What is the maximum possible voltage given by the outlet.

Answer

Since the given voltage is the root mean squared voltage we must multiply the voltage by $\sqrt{2}$ to find the maximum voltage.

$120V \cdot\sqrt{2} = 120V \cdot 1.4= 170V$

We determine that the maximum voltage delivered by the outlet is .

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Question

A neutral conducting sphere rests on an insulating styrofoam block. A student then places his hand on the sphere, maintaining this contact as a negatively charged plastic rod is brought near the sphere without touching. While the plastic rod is still nearby, the student removes his hand. Finally, the rod is also removed. Which of the following best describes the final charge on the sphere?

Answer

The student's hand on the sphere "grounds" it, allowing charges to flow in or out. No charge flows through the styrofoam insulator. When the negatively charged rod is brought nearby, negative charges in the sphere are repelled and exit via the student's hand, giving the sphere a net positive charge. After the student removes his hand, no more charges can flow into or out from the sphere, so the net positive charge remains after the rod is removed. Since this is a conducting sphere, the charges will spread out evenly over the surface.

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Question

How much energy is required by a lightbulb running for minutes?

Answer

Power is given in watts which is equal to $\frac{J}{sec}$

Convert 10 minutes into seconds and then multiply the time in seconds with the number of watts to obtain the joules of energy required by the lightbulb.

$10minutes\cdot\frac{60 seconds}{1 minute}= 600 s$

$40W= 40\frac{J}{sec}$

$40\frac{J}{sec} \cdot 600sec = 24,000J$

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Question

The primary coil of a transformer has 800 turns and an rms voltage of 400V. If an output voltage of 120V is needed, how many turns whould the secondary coil contain?

Answer

Relevant equations:

$\frac{V_s}{V_p}=\frac{N_s}{N_p}$

For the primary coil, we have:

And for the secondary coil:

Plugging these in yields:

$\frac{120}{400}=\frac{N_s}{800}$

$N_s = 240\ \text{turns}$

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Question

An electrical current is flowing through a block. It is observed that the electrical conductivity of the block is $0.417: \frac{\text{Siemens}}{\text{meter}}$ , the length of the block is and the resistance of the block is $3\Omega$ . Which of the following is a valid conclusion about this block?

Answer

To answer this question you need to understand the relationship between electrical conductivity, $\sigma$ , and electrical resistivity, $\rho$ :

$\sigma= \frac{1}{\rho }$

This means that the electrical conductivity is the reciprocal of the electrical resistivity; therefore, the electrical resistivity of this block is:

$\rho = \frac{1}{\sigma }= \frac{1}{0.417\frac{Siemens}{meter}}$

Recall the definition of resistivity:

$\rho = \frac{R\times L}{A}$

Here, is the resistance, is the cross-sectional area, and is the length of the block. The question gives us resistance and length of the block, and we calculated resistivity; therefore, solving for the area of the block gives us:

$A= \frac{\rho \times L}{R}= \frac{2.4\times 5}{3} = 4: m^2$

The cross-sectional area of the block is .

Of the given answer choices, the only valid conclusion is that the block has a square cross-section with a height and width of because this square has an area equal to the cross-sectional area of the block ().

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Question

Consider a conducting rod. Which of the following is true regarding the relationship between the electrical conductivity, the resistance, and the length of the rod?

Answer

The electrical conductivity is the reciprocal of electrical resistivity; therefore, an increase in electrical resistivity will lead to a decrease in electrical conductivity, and vice versa.

Recall the definition of electrical resistivity:

$\rho = \frac{R\times L}{A}$

Here, is resistance, is cross-sectional area, and is the length of the rod. This equation reveals that an increase in resistance and area will increase resistivity, whereas an increase in length will decrease resistivity. Since conductivity is the reciprocal of resistivity, increasing resistance and area will decrease conductivity, whereas increasing the length will increase conductivity.

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