Two point charges `+10muC` and `-15muC` are placed at the vertices A and B of a right angle `Delta^("le")` ABC, such that `AB=0.05m, BC=0.12m`. Calculate the result field intensity at the point C. If `+100muC` of charge is palced at C, then what force will be experienced by the point charge at C?

Explanation

`AC=0.01sqrt(5^(2)+12^(2))=0.01xx13=0.13m`
`cos theta'=0.12/0.13=0.9230`
`cos (180-theta)=-cos theta' =0.9230`

`R=sqrt(P^(2)+Q^(2)+2PQcos theta)`
`tan alpha=(E_(1)sin theta)/(E_(2)+E_(1)cos theta)`
`E=(1/(4 pi epsilon_(0)))q/(x^(2))Vm^(-1)`
`E_(1)=((9xx10^(9))(10xx10^(-6)))/((0.13)^(2))=532.5xx10^(4)Vm^(-1)`
Similarly `E_(2)=((9xx10^(9))(15xx10^(-6)))/((0.12)^(2))=9375xx10^(4)Vm^(-1)`
`E=sqrt(E_(1)^(2)+E_(2)^(2)+2E_(1)E_(2)cos theta)`
`=E=10^(6)sqrt((5.3)^(2)+(9.4)^(2)+2(5.3)(9.4)(-0.923))`
`=10^(6)sqrt(28.09+88.36-91.96)`
`=4.94xx10^(-6)Vm^(-1)` along CD
`tan alpha=(E_(1)sin theta)/(E_(2)+E_(1) cos theta)=(5.3xx10^(6)xx0.05/0.13)/(10^(6)(9.4+5.3xx-0.023))`
`=2.0334/4.5081, alpha=tan^(-1)(0.4521)`
`alpha=24^(@)19^(1)`
but `F=EQ=4.94xx10^(6)xx100xx10^(-6)=4.9xx10^(2)N`