Question

Two charges, Q_{1} and Q_{2}, are separated by
6·cm. The repulsive force between them is 25·N. In each case below,
find the force between them if the **original
situation** is changed as described.

(a) The magnitude of charge Q_{1} is reduced to
Q_{1}/5. N

(b) The distance between the charges is reduced to
2·cm. N

(c) The magnitude of charge Q_{2} is increased to
5Q_{2} and the distance is increased to
30·cm. N

Answer #1

a) The force between charges is proportional to magnitude of charge

Therefore, if magnitude of charge Q1 is reduced to Q1/5, the foece will also be reduced to (1/5)*25=5 N

b)initial distance = 6 cm

new distance = 2 cm

force between two charges is inversely proportional to square of distance between charges

Therefore ,new charge = 25*6^{2}/2^{2} = 225
N

c) force is directly proportional to magnitude of charge and inversely proportional to square of distance between them

Therefore, new force = 25*5*6^{2}/30^{2} = 5
N

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