Question

Solve numerical example.

Four charges of + 6×10^-8 C each are placed at the corners of a square whose sides are 3 cm each. Calculate the resultant force on each charge and show in the direction and a diagram drawn to scale.

Answer 1

Given: q_A = q_B = q_C = q_D = 6 × 10^-8 C, a = 3 cm

Magnitude of force on A due to D is,

F_AD = `1/(4πε_0)"q"^2/"r"_"AD"^2`

= `(9xx10^9xx(6xx10^-8)^2)/(3xx10^-2)^2`

= 3.6 × 10⁻² N

Similarly,

F_AB = 3.6 × 10⁻² N

F_AC = `1/(4πε_0)"q"^2/"r"_"AC"^2`

= `(9xx10^9xx(6xx10^-8)^2)/(3sqrt2xx10^-2)^2`

= 1.8 × 10⁻² N

∴ Resultant force on ‘A’

= F_AD cos 45 + F_AB cos 45 + F_AC

`=(3.6xx10^-2xx1/sqrt2)+(3.6xx10^-2xx1/sqrt2) +1.8xx10^-2`

= 6.89 × 10⁻² N directed along `vec"F"_"AC"`

The resultant force on each charge will be 6.89 × 10⁻² N