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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.
Solution
Given: qA = qB = qC = qD = 6 × 10-8 C, a = 3 cm
Magnitude of force on A due to D is,
FAD = `1/(4πε_0)"q"^2/"r"_"AD"^2`
= `(9xx10^9xx(6xx10^-8)^2)/(3xx10^-2)^2`
= 3.6 × 10−2 N
Similarly,
FAB = 3.6 × 10−2 N
FAC = `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’
= FAD cos 45 + FAB cos 45 + FAC
`=(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
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