Question

Four equal charges each of + 2uc are placed
at the corners of a square of side 50 cm.
Find the force on any one of the charges.​

Answer 1

The force on any one of the charges is 27.56

Explanation:

q₁ = q₂ = q₃ = q₄ = 2 × 10⁻⁶ C

v = 50 cm = 0.5 m

Force acting on one corner is given as:

c = Fca + Fcb + Fcd

On considering x components, we get,

c = Fca cos 0 + Fcb cos 45 + Fcd cos 90

c = Fca + Fcb cos 45 + 0 = Fca + Fcb cos 45

On substituting the values, we get,

$c=\frac{k\left(2 \times 10^{-6}\right)^{2}}{\left(5 \times 10^{-2}\right)^{2}}+\frac{k\left(2 \times 10^{-6}\right)^{2}}{\left(5 \times 10^{-2}\right)^{2}} \frac{1}{2 \sqrt{2}}$

$c =\mathrm{kq}^{2}\left(\frac{1}{25 \times 10^{-4}}+\frac{1}{50 \sqrt{2} \times 10^{-4}}\right)$

$c=\frac{9 \times 10^{9} \times 4 \times 10^{-12}}{24 \times 10^{-4}}\left(1+\frac{1}{2 \sqrt{2}}\right)$

c = 1.44 × 1.35

∴ c = 19.49

Now, the resultant force is given as:

R = √(Fx² + Fy²)

On substituting the values, we get,

R = 19.49√2

∴ R = 27.56