MCAT Physical - Electrostatics and Electrical Fields

1

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?

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

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

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 ().

2

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?

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

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

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 ().

3

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?

Increase distance D by a factor of two.

Decrease distance D by a factor of two.

Decrease the each charge Q by a factor of four.

Increase distance d by a factor of 4.

Explanation

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.

4

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?

Increase distance D by a factor of two.

Decrease distance D by a factor of two.

Decrease the each charge Q by a factor of four.

Increase distance d by a factor of 4.

Explanation

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.

5

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?

The electrical conductivity increases when the length of the rod increases and the resistance of the rod decreases

The electrical conductivity increases when both the length and the resistance of the rod increase

The electrical conductivity increases when both the length and the resistance of the rod decrease

The electrical conductivity increases when the length of the rod decreases and the resistance of the rod increases

Explanation

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.

6

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?

The electrical conductivity increases when the length of the rod increases and the resistance of the rod decreases

The electrical conductivity increases when both the length and the resistance of the rod increase

The electrical conductivity increases when both the length and the resistance of the rod decrease

The electrical conductivity increases when the length of the rod decreases and the resistance of the rod increases

Explanation

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.

7

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, what direction do the electric field lines point?

To the right

To the left

Into the page

Out of the page

Explanation

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.

Vt_question_3

8

A researcher has two rods: rod A and rod B. Rod A has the same geometrical properties as rod B, but has a larger resistance. What can you conclude about the relative rates of charge flow in the two rods?

Rod B has a higher rate of charge flow than rod A because rod B has the higher electrical conductivity

Rod A has a higher rate of charge flow than rod B because rod A has the higher electrical conductivity

Rod A has a higher rate of charge flow than rod B because rod A has the higher electrical resistivity

Rod B has a higher rate of charge flow than rod A because rod B has the higher electrical resistivity

Explanation

The question is asking about the rates of charge flow. Recall that the current is defined as the amount of charge flowing through a point in a given time; therefore, we are looking for the amount of current flowing through the two rods. Current flow is higher for a material with higher electrical conductivity. This means that we need to find the rod with the higher electrical conductivity.

Since there are no easy equations to find conductivity, we need to find the resistivity first. Electrical conductivity is the reciprocal of electrical resistivity (measure of the ability of a material to resist current flow); therefore, an increase in resistivity leads to a decrease in conductivity, and vice versa. Resistivity is defined as:

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

Here, is the resistance, is the cross-sectional area, and is the length of the rod. The question states that the two rods have the same geometrical properties; therefore, the area and the length of the rods are the same. However, the resistance of rod A is higher. This means that the resistivity is higher and conductivity, consequently, is lower for rod A. Since it has a lower conductivity, rod A has a lower charge flow rate.

9

A researcher has two rods: rod A and rod B. Rod A has the same geometrical properties as rod B, but has a larger resistance. What can you conclude about the relative rates of charge flow in the two rods?

Rod B has a higher rate of charge flow than rod A because rod B has the higher electrical conductivity

Rod A has a higher rate of charge flow than rod B because rod A has the higher electrical conductivity

Rod A has a higher rate of charge flow than rod B because rod A has the higher electrical resistivity

Rod B has a higher rate of charge flow than rod A because rod B has the higher electrical resistivity

Explanation

The question is asking about the rates of charge flow. Recall that the current is defined as the amount of charge flowing through a point in a given time; therefore, we are looking for the amount of current flowing through the two rods. Current flow is higher for a material with higher electrical conductivity. This means that we need to find the rod with the higher electrical conductivity.

Since there are no easy equations to find conductivity, we need to find the resistivity first. Electrical conductivity is the reciprocal of electrical resistivity (measure of the ability of a material to resist current flow); therefore, an increase in resistivity leads to a decrease in conductivity, and vice versa. Resistivity is defined as:

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

Here, is the resistance, is the cross-sectional area, and is the length of the rod. The question states that the two rods have the same geometrical properties; therefore, the area and the length of the rods are the same. However, the resistance of rod A is higher. This means that the resistivity is higher and conductivity, consequently, is lower for rod A. Since it has a lower conductivity, rod A has a lower charge flow rate.

10

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?

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

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.

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

Electrostatics and Electrical Fields

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