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MCAT Physical : Electricity and Magnetism

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Example Question #21 : Electricity And Magnetism

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.

Pretext Question_2

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

Possible Answers:

To the left

Out of the page

Into the page

To the right

Correct answer:

To the right

Explanation:

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.

Vt_question_4

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Example Question #4 : Electric Fields

Consider a spherical shell of radius has a charge .

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

Possible Answers:

$\frac{kQ}{9R^2}$

$\frac{kQ}{R}$

$\frac{kQ}{16R^2}$

$\frac{kQ}{4R^2}$

Correct answer:

$\frac{kQ}{9R^2}$

Explanation:

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, r=3R since we are at a distance of from the surface of the shell and therefore 2R+R 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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Example Question #2 : Electrostatics And Electrical Fields

Consider a spherical shell of radius has a charge .

What is the electric field 0.5R away from the center of the shell?

Possible Answers:

$\frac{kQ}{R}$

$\frac{kQ}{0.25R^2}$

$\frac{2kQ}{R^2}$

Correct answer:

Explanation:

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

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Example Question #11 : Electrostatics And Electrical Fields

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

Possible Answers:

The distance was doubled

The distance was halved

The distance was quartered

The distance was quadrupled

The distance was not changed

Correct answer:

The distance was quartered

Explanation:

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_2=4U

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

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

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

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

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Example Question #1 : Electric Potential Energy

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?

Possible Answers:

$-1.7 \times 10^{-20} \textup{J}$

$-2.5 \times 10^{-19} \textup{J}$

$1.5 \times 10^{-18} \textup{J}$

$-1.5 \times 10^{-18} \textup{J}$

Correct answer:

$-1.5 \times 10^{-18} \textup{J}$

Explanation:

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

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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Example Question #13 : Electrostatics And Electrical Fields

Which of the following is not true regarding electric potential?

Possible Answers:

When a positive charge moves from a region of low potential to high potential, the electric field does positive work on the charge

Electric potential can be expressed with units of Volts or Joules per Coulomb

The positive terminal of a battery has higher electric potential than the negative terminal

A negative charge moving from low potential to high potential will accelerate

All of these are true statements

Correct answer:

When a positive charge moves from a region of low potential to high potential, the electric field does positive work on the charge

Explanation:

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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Example Question #14 : Electrostatics And Electrical Fields

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?

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

k=9*10^9N

Possible Answers:

99mJ

9.9J

5.4mJ

720J

7.2J

Correct answer:

9.9J

Explanation:

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

$U_{12} = \frac{kq_1q_2}{r_{12}}= \frac{(9*10^9)(1*10^{-6})(2*10^{-6})}{0.01}=1.8 J$

Charges 1 and 3

$U_{13}=\frac{kq_1q_3}{r_{13}}=\frac{(9*10^9)(1*10^{-6})(3*10^{-6})}{0.01}=2.7J$

Charges 2 and 3

$U_{23} = \frac{kq_2q_3}{r_{23}}=\frac{(9*10^9)(2*10^{-6})(3*10^{-6})}{0.01}=5.4J$

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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Example Question #15 : Electrostatics And Electrical Fields

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

Possible Answers:

170V

120V

220V

100V

Correct answer:

170V

Explanation:

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 170V .

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Example Question #16 : Electrostatics And Electrical Fields

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?

Possible Answers:

$60\ \text{turns}$

$1600\ \text{turns}$

$2700\ \text{turns}$

$400\ \text{turns}$

$240\ \text{turns}$

Correct answer:

$240\ \text{turns}$

Explanation:

Relevant equations:

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

For the primary coil, we have:

N_p = 800

V_p = 400V

And for the secondary coil:

N_s = ?

V_s = 120V

Plugging these in yields:

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

$N_s = 240\ \text{turns}$

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Example Question #17 : Electrostatics And Electrical Fields

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?

Possible Answers:

The block has a circular cross-section that has a radius of

The block has a square cross-section, with a height and width of

The block has a rectangular cross-section with a height of and the width of

The block has a circular cross-section that has a diameter of

Correct answer:

The block has a square cross-section, with a height and width of

Explanation:

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 4m^2 .

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 ( 4m^2 ).

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MCAT Physical : Electricity and Magnetism

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #21 : Electricity And Magnetism

Example Question #4 : Electric Fields

Example Question #2 : Electrostatics And Electrical Fields

Example Question #11 : Electrostatics And Electrical Fields

Example Question #1 : Electric Potential Energy

Example Question #13 : Electrostatics And Electrical Fields

Example Question #14 : Electrostatics And Electrical Fields

Example Question #15 : Electrostatics And Electrical Fields

Example Question #16 : Electrostatics And Electrical Fields

Example Question #17 : Electrostatics And Electrical Fields

All MCAT Physical Resources

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