Calculating the Total Electric Field at the Center of a Square

Objective and Problem Setup

The primary goal of this session is to calculate the total electric field at the center of a square. Four point charges are situated at the vertices of this square, each possessing different magnitudes and polarities (some positive, some negative). The process involves calculating the individual electric field vectors produced by each charge and performing a vector addition to find the net electric field at the center.

Fundamental Concepts and Conventions

To solve this problem, several key physics conventions and formulas regarding electric fields must be established:

1. Direction of the Electric Field

• Positive Point Charge: The electric field vector produced by a positive charge points away from the charge.
• Negative Point Charge: The electric field vector produced by a negative charge points toward the charge.

2. Magnitude of the Electric Field

The magnitude of the electric field ($E$) created by a point charge can be expressed using Coulomb's law. In some contexts, it is written using the permittivity of free space ($\epsilon_0$):

E = \frac{1}{4\times \text{\pi}\times \epsilon_0} \times \frac{|q|}{R^2}

In many textbooks, the constants are grouped into the Coulomb constant ($K$):

$E = \frac{K \times |q|}{R^2}$

• Constant $K$ Value: $K \approx 9 \times 10^9\,\text{N}\cdot\text{m}^2\cdot\text{C}^{-2}$
• Permittivity of Free Space ($\epsilon_0$): This is a constant numerical value representing the ability of a vacuum to permit electric field lines.

Geometric Calculations

For a square with a side length $a = 5\,\text{cm}$, every charge at a vertex is equidistant from the center. We define this distance as $r$.

Calculating the Distance $r$

Using the Pythagorean theorem, where the distance from the center to a corner is the hypotenuse of a right triangle with sides equal to half the square's side ($\frac{a}{2}$):

$r = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2}$

$r = \sqrt{2 \times \left(\frac{a}{2}\right)^2}$

$r = \frac{a}{\sqrt{2}}$

Given the side length $a = 5\,\text{cm}$ ($0.05\,\text{m}$), this value will be used in the denominator of the electric field equations.

Charge Magnitudes and Vector Plotting

The system consists of four charges:

• $Q_1 = +10\,\text{nC}$
• $Q_2 = -20\,\text{nC}$
• $Q_3 = +20\,\text{nC}$
• $Q_4 = -10\,\text{nC}$

Individual Electric Field Vectors at the Center

• $\mathbf{E}_1$ (from $Q_1$): Since $Q_1$ is positive, the field points away from it.
• $\mathbf{E}_3$ (from $Q_3$): Since $Q_3$ is positive ($20\,\text{nC}$), the field points away from it. This vector is twice as long as $\mathbf{E}_1$ because the charge magnitude is doubled while the distance remains the same ($R$).
• $\mathbf{E}_4$ (from $Q_4$): Produced by a negative $-10\,\text{nC}$ charge; the field points toward the charge.
• $\mathbf{E}_2$ (from $Q_2$): Produced by a negative $-20\,\text{nC}$ charge; the field points toward the charge. This vector is twice as long as $\mathbf{E}_4$.

Symmetry and Angle Considerations

Because the geometry is a square, the angle between the diagonal and the horizontal/vertical axes is exactly $45^\circ$.

Vector Magnitude Relationships

By comparing the magnitudes of the charges, we can simplify the vector addition:

1. $|\mathbf{E}_1| = |\mathbf{E}_4|$ (since both charges have a magnitude of $10\,\text{nC}$).
2. $|\mathbf{E}_2| = |\mathbf{E}_3|$ (since both charges have a magnitude of $20\,\text{nC}$).
3. $|\mathbf{E}_2| = |\mathbf{E}_3| = 2 \times |\mathbf{E}_1|$.

By substituting these relationships, we can write all vector magnitudes in terms of $E_1$:

• $|\mathbf{E}_1| = E_1$
• $|\mathbf{E}_4| = E_1$
• $|\mathbf{E}_2| = 2E_1$
• $|\mathbf{E}_3| = 2E_1$

Summation of Vectors and Component Analysis

To find the total electric field ($\mathbf{E}_{\text{total}}$), we break the four vectors into X and Y components.

X-Component Analysis

Due to the symmetry of the square and the arrangement of the magnitudes, the X-components of the field vectors cancel each other out. For every vector pointing right with a specific magnitude, there is a corresponding vector component pointing left that negates it.

Y-Component Analysis

The calculation focuses on the vertical (Y) direction. We assign components based on whether they point up (positive) or down (negative):

• Top Vectors (pointing up): There are two components of magnitude $2E_1 \times \cos(45^\circ)$.
• Bottom Vectors (pointing down): There are two components of magnitude $E_1 \times \cos(45^\circ)$.

Summing these up using vector addition:

$\mathbf{E}_{\text{total}, y} = 2E_1\cos(45^\circ) + 2E_1\cos(45^\circ) - E_1\cos(45^\circ) - E_1\cos(45^\circ)$

$\mathbf{E}_{\text{total}, y} = 4E_1\cos(45^\circ) - 2E_1\cos(45^\circ)$

$\mathbf{E}_{\text{total}, y} = 2E_1\cos(45^\circ)$

This confirms the total electric field points exclusively in the vertical direction ($\mathbf{\hat{j}}$).

Numerical Calculation

Now, we substitute the known values into the finalized expression for total electric field magnitude:

1. Constants:

   • $K = 9 \times 10^9\,\text{N}\cdot\text{m}^2\cdot\text{C}^{-2}$
   • $Q_1 = 10 \times 10^{-9}\,\text{C}$
   • $a = 0.05\,\text{m}$
   • $R^2 = \left(\frac{0.05}{\sqrt{2}}\right)^2$
   • $\cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707$

2. Equation for $E_1$:$E_1 = \frac{K \times Q_1}{R^2}$

3. Total Electric Field equation:$\mathbf{E}_{\text{total}} = 2 \times \left( \frac{K \times Q_1}{R^2} \right) \times \cos(45^\circ)\,\mathbf{\hat{j}}$

4. Substitution:$\mathbf{E}_{\text{total}} = 2 \times \left( \frac{9 \times 10^9 \times 10 \times 10^{-9}}{\left(\frac{0.05}{\sqrt{2}}\right)^2} \right) \times \cos(45^\circ)\,\mathbf{\hat{j}}$

Executing the calculation results in:

$|\mathbf{E}_{\text{total}}| = 1.02 \times 10^5\,\text{N}\cdot\text{C}^{-1}$

Final Result: The total electric field at the center of the square is $1.02 \times 10^5\,\text{N/C}$ pointing in the vertical ($\mathbf{\hat{j}}$) direction.