calculate the force between the two charges when the charges are doubled and distance of separation is halved
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Force between two charges according to Columbs law is
F1 == kq1q2/r²
(k = 1/4πe•)
F when two charges are doubles and distance halved
F 2= k x 4 x (q1q2) ÷ (r/2)²
(The 2² in the double denominator comes to the numerator)
(And 2² = 4)
F2 == 4 x k x 4 x (q1q2)
F2 == 16 x k x q1q2
Hence F2 = 16 times F1
F1 == kq1q2/r²
(k = 1/4πe•)
F when two charges are doubles and distance halved
F 2= k x 4 x (q1q2) ÷ (r/2)²
(The 2² in the double denominator comes to the numerator)
(And 2² = 4)
F2 == 4 x k x 4 x (q1q2)
F2 == 16 x k x q1q2
Hence F2 = 16 times F1
Maryamkhan123:
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