two charges of equal magnitude exact the force of 2N on each other if the direction between them is reducced by 0.5 m then force between them then become 18N and calculate the following (1) initial distance between the charges (2) magnitude of each charges
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F=1/4π€ Qq/r²
it is given that F=2N
2=9×10^9 ×Qq/r²
Qq=2r²/9×10^9
now,RBIs reduced to (r-0.5)m
F=18N
F=9×10^9 ×Qq/(r-0.5)²
putting the value ofQq
18=9×10^9×2r²/(r-0.5)²×9×10^9
18=2r²/(r-0.5)²
9(r-0.5)²=r²
9(r²+0.25-r)=r²
9r²-r²+0.25×9-9r=0
800r²+225-900r=0
800r²-900r+225=0
25(32r²-36r+9)=0
32r²-36r+9=0
32r²-24r-12r+9=0
8r(4r-3)-3(4r-3)=0
r=3/4 or 3/8.
it is given that F=2N
2=9×10^9 ×Qq/r²
Qq=2r²/9×10^9
now,RBIs reduced to (r-0.5)m
F=18N
F=9×10^9 ×Qq/(r-0.5)²
putting the value ofQq
18=9×10^9×2r²/(r-0.5)²×9×10^9
18=2r²/(r-0.5)²
9(r-0.5)²=r²
9(r²+0.25-r)=r²
9r²-r²+0.25×9-9r=0
800r²+225-900r=0
800r²-900r+225=0
25(32r²-36r+9)=0
32r²-36r+9=0
32r²-24r-12r+9=0
8r(4r-3)-3(4r-3)=0
r=3/4 or 3/8.
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