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

Options

A. 0.2 dyne
B. 0.2 newton
C. 2 dyne
D. 0.02 newton

Answer

To find the magnitude of the resultant force on the charge at point D due to the other charges at points A, B, and C, we will follow these steps:

### Step 1: Identify the charges and their positions
- Charge at A: \( q_A = +10^{-7} \, C \)
- Charge at B: \( q_B = -10^{-7} \, C \)
- Charge at C: \( q_C = -2 \times 10^{-7} \, C \)
- Charge at D: \( q_D = +2 \times 10^{-7} \, C \)

The square has a side length of \( 0.05 \, m \).

### Step 2: Calculate the force on charge D due to charge A
The force between two point charges is given by Coulomb's law:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) and \( r \) is the distance between the charges.

The distance between A and D is \( 0.05 \, m \).
\[ F_{AD} = k \frac{|q_A q_D|}{(0.05)^2} = 9 \times 10^9 \frac{|10^{-7} \cdot 2 \times 10^{-7}|}{(0.05)^2} \]
\[ = 9 \times 10^9 \frac{2 \times 10^{-14}}{0.0025} = 9 \times 10^9 \times 8 \times 10^{-12} = 0.072 \, N \]
The direction of this force is away from A since both charges are of opposite signs.

### Step 3: Calculate the force on charge D due to charge B
The distance between B and D is also \( 0.05 \, m \).
\[ F_{BD} = k \frac{|q_B q_D|}{(0.05)^2} = 9 \times 10^9 \frac{|(-10^{-7}) (2 \times 10^{-7})|}{(0.05)^2} \]
\[ = 9 \times 10^9 \frac{2 \times 10^{-14}}{0.0025} = 9 \times 10^9 \times 8 \times 10^{-12} = 0.072 \, N \]
The direction of this force is towards B since both charges are of opposite signs.

### Step 4: Calculate the force on charge D due to charge C
The distance between C and D is \( 0.05 \, m \).
\[ F_{CD} = k \frac{|q_C q_D|}{(0.05)^2} = 9 \times 10^9 \frac{|(-2 \times 10^{-7}) (2 \times 10^{-7})|}{(0.05)^2} \]
\[ = 9 \times 10^9 \frac{4 \times 10^{-14}}{0.0025} = 9 \times 10^9 \times 16 \times 10^{-12} = 0.144 \, N \]
The direction of this force is towards C since both charges are of opposite signs.

### Step 5: Determine the resultant force on charge D
Now we need to consider the directions of the forces:
- \( F_{AD} \) is directed away from A.
- \( F_{BD} \) is directed towards B.
- \( F_{CD} \) is directed towards C.

Since A and C are on the opposite corners of the square, the forces \( F_{AD} \) and \( F_{CD} \) will be in opposite directions, while \( F_{BD} \) will be perpendicular to the line connecting A and C.

Using vector addition:
1. The net force due to A and C:
\[ F_{net, AC} = F_{AD} - F_{CD} = 0.072 - 0.144 = -0.072 \, N \quad (\text{towards C}) \]

2. The resultant force \( F_{resultant} \) can be calculated using Pythagorean theorem since \( F_{BD} \) is perpendicular to \( F_{net, AC} \):
\[ F_{resultant} = \sqrt{(F_{net, AC})^2 + (F_{BD})^2} \]
\[ = \sqrt{(-0.072)^2 + (0.072)^2} = \sqrt{2 \times (0.072)^2} = 0.072 \sqrt{2} \approx 0.102 \, N \]

### Final Result
The magnitude of the resultant force on the charge at D is approximately \( 0.102 \, N \).