D2-2. Electric Field Strength and Superposition

Example 1

Problem: A $+3\text{ C}$ charge and a $-4\text{ C}$ charge are a distance $x$ apart. $P$ is a distance $x$ from the $+3\text{ C}$ charge on the straight line joining the charges.

What is the magnitude of the electric field strength at $P$?

Solution:

Point $P$ is a distance $x$ from the $+3\text{ C}$ charge. The charges themselves are $x$ apart. Based on the standard configuration for this question, $P$ is to the left of the $+3\text{ C}$ charge, making it a total distance of $2x$ from the $-4\text{ C}$ charge.
The electric field from $+3\text{ C}$ points Left (away from positive) with magnitude $$ E_1 = k \frac{3}{x^2} $$ The electric field from $-4\text{ C}$ points Right (towards negative) with magnitude $$ E_2 = k \frac{4}{(2x)^2} = k \frac{4}{4x^2} = k \frac{1}{x^2} $$ Since they are in opposite directions, we subtract them to find the net field magnitude:
$$ E_{net} = E_1 - E_2 = \frac{3k}{x^2} - \frac{1k}{x^2} = \mathbf{\frac{2k}{x^2}} $$

Example 2

Problem: A negatively charged particle in a uniform gravitational field is positioned mid-way between two charged conducting plates.

The potential difference between the plates is adjusted until the particle is held at rest relative to the plates. What change will cause the particle to accelerate downwards relative to the plates?
A. Decreasing the charge on the particle
B. Decreasing the separation of the plates
C. Increasing the length of the plates
D. Increasing the potential difference between the plates

Solution:

If the particle is at rest, the downward gravitational force ($F_g = mg$) exactly equals the upward electrostatic force ($F_e = qE = qV/d$). To make it accelerate downwards, we must make the upward electrostatic force ($F_e$) weaker.
Looking at the options:
- Decreasing the separation ($d$) increases $F_e$ (particle moves up).
- Increasing potential difference ($V$) increases $F_e$ (particle moves up).
- Decreasing the charge ($q$) on the particle reduces $F_e$, making the downward gravity relatively stronger. Therefore, the particle will accelerate downwards. Answer is A.

Example 3

Problem: Three point charges of equal magnitude are placed at the vertices of an equilateral triangle. The signs of the charges are shown. Point $P$ is equidistant from the vertices of the triangle. What is the direction of the resultant electric field at $P$?

Solution:

Assume an imaginary positive test charge at center $P$.
1. The top-left vertex is Positive. It pushes $P$ down and to the right.
2. The top-right vertex is Positive. It pushes $P$ down and to the left.
3. The bottom vertex is Negative. It pulls $P$ straight down.
The horizontal (left and right) pushes from the two top charges cancel each other out completely. However, all three charges contribute a downward vertical force. Therefore, the resultant electric field points straight down (in the direction of Vector C).

Example 4

Problem: Two charges, $+Q$ and $-Q$, are placed as shown.

What is the magnitude of the electric field strength, in descending order, at points $X$, $Y$ and $Z$?

Solution:

The total electric field strength ($E$) at each point is the vector sum of the individual fields from the $+Q$ and $-Q$ charges ($E = E_{+Q} + E_{-Q}$). The magnitude of each field follows the inverse-square law ($E \propto 1/r^2$). The field from $+Q$ always points away from the charge, and the field from $-Q$ always points towards the charge.

• Point Z (Maximum Magnitude): Both Z and Y are equidistant (1 cm) from the strong $+Q$ charge. At point Z, the electric field vectors from $+Q$ and $-Q$ both point to the right. Because they add together in the same direction, this results in the strongest net field ($E_Z$ is max).
• Point Y (Intermediate Magnitude): At point Y, the strong field from $+Q$ points left, while the weaker field from $-Q$ points right. Because these vectors oppose each other, they partially cancel out, making the net field weaker than at Z.
• Point X (Lowest Magnitude): This point is the furthest outside, 2 cm from $+Q$. The field from $+Q$ (pointing left) is significantly weaker at this distance, and it is further opposed by the field from $-Q$ (pointing right), leading to the lowest total magnitude.

Conclusion: Therefore, the descending magnitude order is $\mathbf{Z > Y > X}$.

Example 5

1. Define electric field strength at a point in an electric field.

2. Six point charges of equal magnitude $Q$ are held at the corners of a hexagon with the signs of the charges as shown. Each side of the hexagon has a length $a$. $P$ is at the center of the hexagon.

   1. Show, using Coulomb’s law, that the magnitude of the electric field strength at point $P$ due to one of the point charges is $kQ/a^2$.
   2. On the diagram, draw arrows to represent the direction of the field at $P$ due to point charge $A$ (label this direction $A$) and point charge $B$ (label this direction $B$).
   3. The magnitude of $Q$ is $3.2\text{ }\mu\text{C}$ and length $a$ is $0.15\text{ m}$. Determine the magnitude and the direction of the electric field strength at point $P$ due to all six charges.

Solution:

1. Electric field strength is the electrostatic force exerted per unit positive test charge ($E = F/q$).
2. 1. By geometry, a regular hexagon is composed of six equilateral triangles. Therefore, the distance from any corner to the center $P$ is exactly $a$. Using Coulomb's Law, the field from a single point charge at a distance $a$ is: $$E = \frac{kQ}{a^2}$$
   2. Point charge $A$ (top-left) is Positive, so its electric field vector at $P$ points away from $A$ (down and to the right). Point charge $B$ (bottom-left) is Negative, so its electric field vector at $P$ points towards $B$ (down and to the left).

   3. To find the net electric field, we can evaluate the pairs of charges on opposite corners of the hexagon:

      • The Left (+) and Right (+) charges create equal and opposite fields that push against each other and completely cancel out.
      • The Bottom-Left (-) and Top-Right (-) charges create equal and opposite fields pulling towards themselves that also completely cancel out.
      • The Top-Left (+) and Bottom-Right (-) charges both produce fields pointing in the same direction (down and to the right). These two vectors add together to give $2E$.

      Therefore, the total field at $P$ is simply:

      $$E_{net} = 2 \times \frac{kQ}{a^2} = 2 \times \frac{(8.99 \times 10^9) \times (3.2 \times 10^{-6})}{(0.15)^2} = \mathbf{2.56 \times 10^6 \text{ N C}^{-1}}$$

      The net direction is down and to the right (towards the bottom-right charge).

Example 6

Problem: The diagram shows electrostatic field lines. The source of the field is not shown. At which position in the field would a negative point charge experience the greatest force to the right?

Solution:

The magnitude of the force is greatest where the electric field is strongest. Electric field strength is visually represented by the density (closeness) of the field lines. Therefore, the charge must be placed where the lines are bunched tightly together. A negative charge experiences a force in the opposite direction of the electric field. Because the field lines point to the left, a negative charge will experience a force to the right at all marked positions. In the given diagram, the field lines are drawn significantly closer together at position D than at positions A, B, or C, indicating the strongest electric field is located there. Therefore, the negative charge would experience the greatest force to the right at position D.