Relationships Between Force, Field, Energy, Potential: Videos & Practice Problems

Understanding electric forces, fields, potential energy, and potential is crucial in physics. The electric field (E) exerts a force (F) on a charge (q) as described by the equation F^2 = qE. Similarly, potential energy (U) relates to potential (V) through U = qV. The relationship between potential energy and force is given by -ΔU = FΔr, while potential difference (voltage) relates to electric field as -ΔV = EΔr.

Relationships Between Force, Field, Energy, Potential

In the study of electricity, understanding the relationships between electric forces, electric fields, potential energies, and electric potentials is crucial. These concepts are interconnected and can be summarized effectively through key equations and definitions.

The electric field (E) is a force field created by a charge (Q) that exerts a force (F) on another charge (q). The relationship between the electric force and the electric field is expressed by the formula:

$$ F = q \cdot E $$

Here, the force experienced by the charge q is directly proportional to the electric field produced by charge Q. It’s important to note that different notations (such as Q and q) may be used in various contexts, but the underlying principles remain the same.

Additionally, a charge also generates an electric potential (V), which indicates the potential energy (U) per unit charge. The relationship between potential energy and electric potential is given by:

$$ U = q \cdot V $$

To connect these concepts further, the change in potential energy (ΔU) can be related to the work done by the electric force over a distance (Δr or Δx), expressed as:

$$ -\Delta U = F \cdot \Delta r $$

Similarly, the relationship between electric field and electric potential difference (ΔV) is described by:

$$ -\Delta V = E \cdot \Delta r $$

In these equations, Δr represents the distance over which the force acts, and it can be denoted as Δx, Δd, or other variations depending on the context. The term ΔV specifically refers to the voltage, which is the potential difference between two points in an electric field.

Understanding these relationships allows for a comprehensive grasp of how electric forces and fields interact with charges, providing a solid foundation for further studies in electromagnetism. This synthesis of concepts is essential for success in examinations and practical applications in physics.

Potential at Center of Charges in a Square

To determine the electric potential at the center of a square arrangement of charges, we start by recalling the formula for electric potential, which is given by:

$V^{=}\frac{kq}{r}$

Here, k is Coulomb's constant, approximately equal to \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\), q is the charge, and r is the distance from the charge to the point of interest.

In this scenario, we need to find the total potential at the center of the square formed by the charges. The first step is to calculate the distance from the center to each corner of the square. If the side length of the square is \(5 \, \text{mm}\), then the distance to the center can be determined using the Pythagorean theorem. By dividing the square into two right triangles, we find that:

$r=\; \backslash sqrt\{(2.5\; \backslash ,\; \backslash text\{mm\})^2\; +\; (2.5\; \backslash ,\; \backslash text\{mm\})^2\}\; =\; 3.54\; \backslash ,\; \backslash text\{mm\}\; =\; 3.54\; \backslash times\; 10^\{-3\}\; \backslash ,\; \backslash text\{m\}$

Next, we can calculate the total potential at the center by summing the potentials contributed by each charge. Since electric potential is a scalar quantity, we can simply add the potentials from each charge:

$V\_\{\backslash text\{total\}\}\; =\; \backslash sum\; \backslash frac\{k\; \backslash cdot\; q\_i\}\{r\}$

In this case, the charges are \(2 \, \text{nC}\), \(-3 \, \text{nC}\), \(1 \, \text{nC}\), and \(-1.5 \, \text{nC}\). The total potential can be simplified by factoring out \(k/r\):

$V\_\{\backslash text\{total\}\}\; =\; \backslash frac\{k\}\{r\}\; \backslash sum\; q\_i$

Calculating the sum of the charges:

$2\; \backslash ,\; \backslash text\{nC\}\; -\; 3\; \backslash ,\; \backslash text\{nC\}\; +\; 1\; \backslash ,\; \backslash text\{nC\}\; -\; 1.5\; \backslash ,\; \backslash text\{nC\}\; =\; -1.5\; \backslash ,\; \backslash text\{nC\}$

Now substituting the values into the potential formula:

$V\_\{\backslash text\{total\}\}\; =\; \backslash frac\{8.99\; \backslash times\; 10^9\; \backslash ,\; \backslash text\{N\; m\}^2/\backslash text\{C\}^2\}\{3.54\; \backslash times\; 10^\{-3\}\; \backslash ,\; \backslash text\{m\}\}\; \backslash cdot\; (-1.5\; \backslash times\; 10^\{-9\}\; \backslash ,\; \backslash text\{C\})$

Calculating this gives:

$V\_\{\backslash text\{total\}\}\; \backslash approx\; -3.81\; \backslash times\; 10^3\; \backslash ,\; \backslash text\{V\}$

This negative value indicates that the potential at the center is lower than the reference potential, which is an important aspect to consider in electrostatics. Understanding how to simplify the calculations by factoring out common terms can significantly streamline the process of finding electric potentials in complex charge arrangements.

A -2C charge lies at rest. (a) What is the potential difference between point A, which is 1.5m from the charge, and point B, which is 4m from the charge? (b) What would the work on a 4C charge be to move it from A to B?

• A. (a) −6.9×10⁹ V; (b) +2.7×10¹⁰ J
• B. (a) +6.9×10⁹ V; (b) −2.7×10¹⁰ J
• C. (a) −7.5×10⁹ V; (b) +3.0×10¹⁰ J
• D. (a) +7.5×10⁹ V; (b) −3.0×10¹⁰ J
• E. (a) −1.6×10¹⁰ V; (b) +6.4×10¹⁰ J
• F. (a) +1.6×10¹⁰ V; (b) −6.4×10¹⁰ J

Answer: (D)

Potential Difference Between Two Charges

In this scenario, we are analyzing the electric potential created by two point charges: a positive charge of 5 nanocoulombs and a negative charge of 3 nanocoulombs, positioned 6 millimeters apart. To find the electric potential at a point (VA) located halfway between these charges, we can utilize the principle that electric potential (V) due to a point charge is given by the formula:

$$ V = k \frac{q}{r} $$

where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( q \) is the charge, and \( r \) is the distance from the charge to the point of interest. Since point A is equidistant from both charges (3 millimeters), we can simplify the calculation by summing the potentials from both charges:

$$ V_A = k \left( \frac{q_1}{r_A} + \frac{q_2}{r_A} \right) $$

Given that both distances are the same, we can factor out the common terms:

$$ V_A = \frac{k}{r_A} (q_1 + q_2) $$

Substituting the values, we convert 3 millimeters to meters (0.003 m) and calculate:

$$ V_A = \frac{8.99 \times 10^9}{0.003} \times (5 \times 10^{-9} - 3 \times 10^{-9}) $$

This results in a potential of approximately \( 6 \times 10^3 \, \text{V} \) at point A.

Next, we consider point B, which is also halfway between the charges but positioned 4 millimeters above the line connecting the charges. To find the potential at point B (VB), we first determine the distance from point B to each charge using the Pythagorean theorem. The distances to both charges are equal, forming a 3-4-5 triangle, giving us a distance of 5 millimeters (0.005 m) to each charge.

Using the same potential formula:

$$ V_B = \frac{k}{r_B} (q_1 + q_2) $$

Substituting the values, we find:

$$ V_B = \frac{8.99 \times 10^9}{0.005} \times (5 \times 10^{-9} - 3 \times 10^{-9}) $$

This results in a potential of approximately \( 3.6 \times 10^3 \, \text{V} \) at point B.

Finally, to calculate the work done on a 1 nanocoulomb charge moving from point A to point B, we use the work-energy principle, which states that the work (W) done on a charge moving through a potential difference (ΔV) is given by:

$$ W = -q \Delta V $$

where \( \Delta V = V_B - V_A \). Substituting the values:

$$ W = -1 \times 10^{-9} \times (3.6 \times 10^3 - 6 \times 10^3) $$

Calculating this gives a work done of approximately \( 2.4 \times 10^{-6} \, \text{J} \). This negative value indicates that work is done against the electric field when moving the charge from point A to point B.

Distance to Stop a Point Charge

In this scenario, we are analyzing the motion of a charged particle influenced by an electric force. We have a positive charge of 3 microcoulombs (µC) with a mass of 5 grams (0.005 kg) moving away from a negative charge of 5 µC. The initial speed of the positive charge is 20 meters per second, and the initial distance between the two charges is 5 centimeters (0.05 meters). The goal is to determine how far the positive charge can travel before it comes to a stop due to the attractive force exerted by the negative charge.

As the positive charge moves away, the electric force, which is attractive, will eventually decelerate it until it stops. To find the distance it travels before stopping, we can apply the principle of conservation of energy. The total mechanical energy in the system remains constant, meaning the initial kinetic energy and initial potential energy will equal the final kinetic energy and final potential energy.

The conservation of energy can be expressed with the formula:

\[ \frac{1}{2} mv_0^2 + U_i = U_f + \frac{1}{2} mv_f^2 \]

Where:

• \(m\) is the mass of the charge (0.005 kg),
• \(v_0\) is the initial velocity (20 m/s),
• \(U_i\) is the initial electric potential energy,
• \(U_f\) is the final electric potential energy,
• \(v_f\) is the final velocity (0 m/s when the charge stops).

The initial kinetic energy can be calculated as:

\[ U_i = k \frac{q_1 q_2}{r_i} \]

And the final potential energy is:

\[ U_f = k \frac{q_1 q_2}{r_f} \]

Here, \(k\) is Coulomb's constant (\(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\)), \(q_1\) and \(q_2\) are the magnitudes of the charges (3 µC and 5 µC), and \(r_i\) and \(r_f\) are the initial and final distances between the charges.

Substituting the known values into the energy conservation equation allows us to solve for the final distance \(r_f\). After calculating the initial kinetic energy and potential energy, we find that:

\[ \frac{1}{2} (0.005) (20^2) + k \frac{(3 \times 10^{-6})(-5 \times 10^{-6})}{0.05} = k \frac{(3 \times 10^{-6})(-5 \times 10^{-6})}{r_f} \]

After performing the calculations, we find that the final distance \(r_f\) is approximately 0.0793 meters (or 7.93 centimeters). To determine how far the charge travels before stopping, we subtract the initial distance from the final distance:

\[ x = r_f - r_i = 7.93 \, \text{cm} - 5 \, \text{cm} = 2.93 \, \text{cm} \]

Thus, the positive charge travels approximately 2.93 centimeters before coming to a stop. Understanding energy conservation in electric fields is crucial for solving similar problems in electrostatics.

Here’s what students ask on this topic:

What is the relationship between electric force and electric field?

The relationship between electric force (F) and electric field (E) is given by the equation:

$F=qE$

Here, F is the force experienced by a charge q in an electric field E. This means that the force on a charge is directly proportional to the magnitude of the charge and the strength of the electric field. If you know the electric field and the charge, you can easily calculate the force acting on the charge.

How is potential energy related to electric potential?

Potential energy (U) is related to electric potential (V) through the equation:

$U=qV$

In this equation, U is the potential energy, q is the charge, and V is the electric potential. This relationship indicates that the potential energy of a charge in an electric field is the product of the charge and the electric potential at that point. It helps in understanding how much energy a charge has due to its position in the electric field.

What is the formula for the relationship between potential energy difference and force?

The relationship between potential energy difference (ΔU) and force (F) is given by:

$-\Delta U=F\Delta r$

Here, ΔU is the change in potential energy, F is the force, and Δr is the displacement. This equation shows that the negative change in potential energy is equal to the work done by the force over a distance. It is crucial for understanding how energy is transferred in an electric field.

How does potential difference relate to the electric field?

The relationship between potential difference (ΔV) and the electric field (E) is given by:

$-\Delta V=E\Delta r$

In this equation, ΔV is the potential difference, E is the electric field, and Δr is the displacement. This relationship indicates that the negative of the potential difference is equal to the electric field multiplied by the distance over which the field is applied. It helps in understanding how the electric field affects the potential difference between two points.

What is the significance of the voltage in electric fields?

Voltage, or electric potential difference (ΔV), is significant in electric fields because it represents the work needed to move a charge between two points in the field. It is defined as:

$\Delta V=V$_final - V_initial

Voltage is crucial for understanding how much energy per unit charge is available to do work. It is a key concept in circuits and electrostatics, influencing how charges move and how much energy is stored in electric fields.