Two point charges experience an attractive force of 8 N when they are separated by 1 m. What is the magnitude of the force they experience when their separation is 0.5 m?
electrostatic force is inversely proportional to the square of distance,hence,
F2/F1 = (R1/R2)2
F2=F1 *(R1/R2)2
F2= 8*(1/0.5)2
F2= 32 N
SOLUTION :
F is inversely proportional to squared distance, d^2
=> F = k / d^2
=> k = F d^2
When F = 8 N and d = 1 m :
=> k = 8 * (1)^2 = 8 N-m^2
So, when distance is 0.5 m :
F2 = k / (0.5)^2 = 8/ (0.5)^2 = 32 N (ANSWER).
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