Question 41: Electrostatics & 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 F
1
. If they are brought in contact and separated to the same distance, they experience force F
2
. The ratio of F
1
to F
2
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

Electrostatics & Capacitance

Two small conducting spheres of equal radius have charges +20 \mu C and -40 \muC respectively and placed at a distance R from each other experience for F 1 . If they are brought in contact and separated to the same distance, they experience force F 2 . The ratio of F 1 to F 2 is

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