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Question
Solve numerical example.
Three equal charges of 10×10-8 C respectively, each located at the corners of a right triangle whose sides are 15 cm, 20 cm, and 25cm respectively. Find the force exerted on the charge located at the 90° angle.
Solution
Given: qA = qB = qC = 10 × 10-8 C
Force on B due to A,
`vec"F"_"BA"=1/(4πε_0)("q"_"A""q"_"B")/(("r"_"AB"^2))`
= `9 × 10^9xx((10xx10^-8)^2)/(20xx10^-2)^2`
= 2.25 × 10−3 N
Force on B due to C,
`vec"F"_"BC"=1/(4πε_0)("q"_"C""q"_"B")/(("r"_"BC"^2))`
= `9xx10^9xx((10xx10^-8)^2)/(15xx10^-2)^2`
= 4 × 10−3 N
∴ Resultant force on point B,
|FB| = `sqrt("F"_"BA"^2+"F"_"BC"^2+2"F"_"BA"."F"_"BC"cos 90)`
= `sqrt((2.25xx10^-3)^2+(4xx10^-3)^2)`
= 4.589 × 10−3 N
Force exerted on charge at point B is 4.589 × 10−3 N.
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