Question

What is the force F on a 1 nC charge placed in the middle of a square with side 1 cm due to the charges of -2nC placed at the four square corners?

Answer 1

The total force on the charge placed in the middle will be zero (0).

Step-by-step explanation:

The force on the charge in the middle of a square due to one of the charges placed at the square corner is given by:

`F = (KQq_(1))/(d^(2))`

Where:

Q: is the charge in the middle = 1 nC

`q_(1) = q_(2) = q_(3) = q_(4)` are the charge placed at the 4 square corners = -2 nC

The distance between the charge Q and q₁, q₂, q₃, and q₄ is:

`d = \frac{\sqrt{2a^(2)}}{2} = \frac{\sqrt{2(1 cm)^(2)}}{2} = 0.707 cm = 7.07 \cdot 10^(-3) m`

Now, the magnitude of the force is:

`|F_{Qq_(1)}| = (KQq_(1))/(d^(2)) = (9\cdot 10^(9)*1 \cdot 10^(-9) C*2 \cdot 10^(-9) C)/((7.07 \cdot 10^(-3) m)^(2)) = 3.60 \cdot 10^(-4) N = |F_{Qq_(2)}| = |F_{Qq_(3)}| = |F_{Qq_(4)}|`

Since the magnitude of the force on the charge placed in the middle of the square is the same due to the 4 charges placed in the square corners (because the charge of q₁, q₂, q₃, and q₄ charges is the same) and due to the geometry, the vectorial sum of the total force on the charge placed in the middle will be zero (0).

Therefore, the total force on the charge placed in the middle will be zero (0).

I hope it helps you!