Question 42: Electrostats & Capacitance

Question

Two small conducting spheres of equal radius have charges +20 $$\mu$$ C and $$-$$40 $$\mu$$C respectively and placed at a distance R from each other experience for F1. If they are brought in contact and separated to the same distance, they experience force F2. The ratio of F1 to F2 is

Accepted Answer

Given, two small conducting spheres of equal radius (R).
Charge on first sphere, q
1
= +20 $$\mu$$C
Charge on second sphere, q
2
= $$-$$40 $$\mu$$C
$$	herefore$$ $${F_1} = {{k( + 20)( - 40)} \over {{R^2}}} = {{ - k(800)} \over {{R^2}}}$$ ..... (i)
The spheres have equal radii, so their capacities will be same. Now, if they are brought in contact, then the net charge,
q
net
= (+20 $$-$$40) $$\mu$$C = $$-$$20 $$\mu$$C
So, each sphere will have charge, $$q = {{{q_{net}}} \over 2} =  - 10$$ $$\mu$$C
Now, $${F_2} = {{k(q)(q)} \over {{R^2}}} = {{k( - 10)( - 10)} \over {{R^2}}} = {{k(100)} \over {{R^2}}}$$ ..... (ii)
Ratio, $${{{F_1}} \over {{F_2}}} = - {{800} \over {100}} =  - 8:1$$
Thus, ratio of F
1
to F
2
= $$-$$8 : 1

Answer

1 : 4

Answer

+8 : 1

Answer

$$-$$8 : 1

Answer

1 : 8

Electrostats & Capacitance

Two small conducting spheres of equal radius have charges +20 $\mu C and -40 \mu$C respectively and placed at a distance R from each other experience for F1. If they are brought in contact and separated to the same distance, they experience force F2. The ratio of F1 to F2 is

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