Four point charge of +10^-7, -10^-7, +2?10^-7, -2?10^-7 are placed respectively at the corners A, B, C, D of a 0.05 m square. Find the magnitude of the resultant force on the charge at D.

To find the resultant force on the charge at point D due to the other three charges placed at points A, B, and C, we need to apply Coulomb's Law. This law states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Let's break this down step by step.

Understanding the Setup

We have four point charges located at the corners of a square with a side length of 0.05 m:

• Charge at A: +10^-7 C
• Charge at B: -10^-7 C
• Charge at C: +2 × 10^-7 C
• Charge at D: -2 × 10^-7 C (the charge we are focusing on)

Calculating Forces from Each Charge

We will calculate the force on the charge at D from each of the other three charges separately and then find the resultant force by vector addition.

Force from Charge A

The distance from D to A is the diagonal of the square, which can be calculated using the Pythagorean theorem:

Distance (DA) = √(0.05^2 + 0.05^2) = √(0.0025 + 0.0025) = √0.005 = 0.07071 m.

Using Coulomb's Law:

F_AD = k * |q_A * q_D| / r_AD²

Where:

• k = 8.99 × 10^9 N m²/C² (Coulomb's constant)
• q_A = +10^-7 C
• q_D = -2 × 10^-7 C
• r_AD = 0.07071 m

Substituting the values:

F_AD = (8.99 × 10^9) * (10^-7 * 2 × 10^-7) / (0.07071)²

Calculating this gives:

F_AD ≈ 0.0051 N

This force is attractive since the charges are of opposite signs, pulling D towards A.

Force from Charge B

The distance from D to B is simply the side of the square, which is 0.05 m.

Using Coulomb's Law again:

F_BD = k * |q_B * q_D| / r_BD²

Where:

• q_B = -10^-7 C

Substituting the values:

F_BD = (8.99 × 10^9) * (10^-7 * 2 × 10^-7) / (0.05)²

Calculating this gives:

F_BD ≈ 0.0719 N

This force is also attractive, pulling D towards B.

Force from Charge C

The distance from D to C is again the side of the square, which is 0.05 m.

Using Coulomb's Law:

F_CD = k * |q_C * q_D| / r_CD²

Where:

• q_C = +2 × 10^-7 C

Substituting the values:

F_CD = (8.99 × 10^9) * (2 × 10^-7 * 2 × 10^-7) / (0.05)²

Calculating this gives:

F_CD ≈ 0.2876 N

This force is repulsive, pushing D away from C.

Finding the Resultant Force

Now we need to consider the directions of these forces:

• F_AD is directed towards A (diagonal direction).
• F_BD is directed towards B (horizontal direction).
• F_CD is directed away from C (horizontal direction).

To find the resultant force, we can break down the forces into their components. The diagonal force F_AD can be split into horizontal and vertical components:

F_AD,x = F_AD * cos(45°) = 0.0051 * (√2/2) ≈ 0.0036 N

F_AD,y = F_AD * sin(45°) = 0.0051 * (√2/2) ≈ 0.0036 N

Now, we can sum the forces in the x and y directions:

• In the x-direction: F_x = F_BD - F_CD + F_AD,x
• In the y-direction: F_y = F_AD,y

Calculating these:

F_x = 0.0719 - 0.2876 + 0.0036 ≈ -0.2121 N

F_y = 0.0036 N