Electric Field Calculation at the Lower Right Corner of a Square with Charges
In this article, we will explore the calculation of the electric field at the lower right corner of a square when three charges, q1 4 x 10^-6 C, q2 -2 x 10^-6 C, and q3 5 x 10^-6 C, are placed at the corners of a square with side lengths of 0.30 meters. The positions of the charges are as follows: q1 at the upper left, q2 at the upper right, and q3 at the lower left. We will calculate the electric field at the lower right corner using vector addition and trigonometry.
Understanding the Problem
The charges form a square, and we need to find the electric field at the fourth corner, specifically at the lower right corner. The angles of the electric field vectors due to each charge are as follows: the field from q1 is at 315°, the field from q2 is at 90°, and the field from q3 is at 0°. We will apply the formula for the electric field and break down the components to find the resultant field.
Electric Field Contributions
The electric field due to a point charge q at a distance r is given by E k * q / r2, where k is Coulomb's constant. For a side length of 0.30 meters, we need to adjust for the diagonal distance in the calculations.
Calculation Steps
The first step is to calculate the electric field contributions from each charge. We will use the formula E k * q / r2 and adjust for the angle and distance.
Field from Charge q1
The electric field from q1 at 315° can be broken into horizontal and vertical components. The horizontal component is E_x1 k * q1 / d2 * cos(315) and the vertical component is E_y1 k * q1 / d2 * sin(315).
Field from Charge q2
The electric field from q2 is at 90°, so its horizontal component is zero and its vertical component is E_y2 k * q2 / d2 * sin(90).
Field from Charge q3
The electric field from q3 at 0° is purely horizontal, so its horizontal component is E_x3 k * q3 / d2 * cos(0) and its vertical component is zero.
Summing the Components
Next, we sum the horizontal and vertical components to find the resultant electric field at the lower right corner.
Horizontal Components
The horizontal components contribute to the resultant field in the following way: E_x 450/9 * 10^-6 * cos(315) 5100/9 * 10^-6 * cos(90) 100/9 * 10^-6 * cos(0). Simplifying this, we get the horizontal component as 640534 N/C.
Vertical Components
The vertical components contribute in a similar manner: E_y 450/9 * 10^-6 * sin(315) 2100/9 * 10^-6 * sin(90) 100/9 * 10^-6 * sin(0). Simplifying this, we get the vertical component as 58498 N/C.
Resultant Electric Field
The resultant electric field is the vector sum of the horizontal and vertical components. Using the Pythagorean theorem, the magnitude of the resultant electric field is:
ER √(E_x2 E_y2) √(6405342 584982) 643200 N/C
The direction of the electric field is given by the arctangent of the vertical component divided by the horizontal component:
θ arctan(58498 / 640534) ≈ 52° north of east
Therefore, the electric field at the lower right corner is 640 kN/C at 52° north of east, with two significant figures.
Conclusion
In conclusion, the electric field at the lower right corner of the square, given the specific positions and magnitudes of the charges, has been calculated and is presented in a step-by-step manner. This calculation demonstrates the application of vector addition and trigonometry in the context of electric field calculations.
Additional Resources
For further reading, you may want to explore the following topics:
Electric fields and their properties Vector addition in physics Applications of trigonometry in physics