To be given in question
Q1=7.5μc
Q2=7.2μc
Q3=-12.9μc
To be asked in question
Electric force on x-axis
We know that
F13=kr2q1q3 →(1)
F23=kr2q2q3→(2)
F=F132+F232+2F13F23cosθ
θ=90°
Put θ value
F=F132+F232→(3)
equation (1)put values
F13=−r1329×109×7.5×10−6×12.9×10−6→(4)
F13=−r1320.87075→(4)
F23=−r2329×109×7.2×10−6×12.9×10−6
F23=r232−0.83592→(5)
equation (3); equation (4)and eqution (5) we can written as
F=(−r1320.87075)2+(r232−0.83592)2→(6)
equation (6)put value r13and r23
and find the resultant force
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