Question

Two small conducting spheres of equal radius have charges +10C and −20C respectively and placed at a distance R from each other experience force 1F . If they are brought in contact and separated to the same distance, they experience force 2 F . The ratio of 1 F to 2 F is??

Answer 1

Answer:

F 1 = 9×10^9 200×10^-6%R^2 and F 2 = -9×10^9 × 10^-6% R^2 now divide F1 and F2 then you will get 4 the option ; CONCEPT behind it is when two charges are attached then they will get same charge as -20+10= -10 C

2.9

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Answer 2

Answer:

F1:F2 = -4:1

Explanation:

Given, Two small conducting spheres of equal radius have charges +10C and −20C respectively and placed at a distance R from each other experience force F1

F1 = - 200/4π€R² (attractive)

Now, When they are brought to contact, flow of charges from the sphere takes place untill both the spheres have the same potential

V1 = V2

q1/r1 = q2/r2. [r1 = r2]

q1 = q2

Hence the charge on both the spheres become equal after contact.

Q = 10C - 20C

Q = -10C = q1 + q2

q1 = q2 = -5C

So, F2 = 25/4π€R² (repulsive)

F1/F2 = -100/25

F1/F2 = -4/1