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The force between two charge when seprate by a distance of 50cm in air is 40 newton what will be the force between them if the distance becomes 25?

The force between two charge when seprate by a distance of 50cm in air is 40 newton what will be the force between them if the distance becomes 25?

Grade:12

5 Answers

Maxx
15 Points
4 years ago
158.4N..................m..................,..................................,.........!!!!!!!!!!!!!!!!!!!!!!!!!!!!
DIWAKAR PANDEY
168 Points
4 years ago
Dear, As given is F=kq1q2/r×r= Here we had changed r only So F is inversely proportional to r square. So F1/F2= 25×25/50×50 ( Cancel 50 with 25) F1=40(given) 40/F2=1/4 So F2=160
shubham singh
13 Points
4 years ago
160 N is answer.....................mmm................................................................mmmmmmm.............................
Shailendra Kumar Sharma
188 Points
4 years ago
F=Kq1q2/r2
so F1/F2=(r2)^2/(r1)^2
So F50 / F25= (25*25)/(50*50) =1/4
so the force at 25 cm will be 4 times of that at 50
so when distance is 25 the force will be 160N
Divya
24 Points
3 years ago
As the charges are same so they will cancel each other.Therefore,F1/F2=25×25/50×5040/F2=1/4F2=40×4=160N

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