if the distance between two equal point charges is doubled and their individual charges are also doubled what wouid happen to the force b/w them?
Answers
Answered by
109
F between them remains same
F= k q1 x q2 / r²
now if we doubled
F= k 2q1 x 2q2 / 4r²
F= 4 k q1 x q2 / 4 r²
F= k q1 x q2 / r²
F= k q1 x q2 / r²
now if we doubled
F= k 2q1 x 2q2 / 4r²
F= 4 k q1 x q2 / 4 r²
F= k q1 x q2 / r²
Answered by
14
f will be same
f = k q1q2/r^2
f=k 2(q1q2)/r^2
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