Two conducting spheres of radii 3 cm and 1 cm are separated by a distance of 10 cm in free space. If the spheres are charged to same potential of 10 V each, the force of repulsion between them is

Question

Two conducting spheres of radii 3 cm and 1 cm are separated by a distance of 10 cm in free space. If the spheres are charged to same potential of 10 V each, the force of repulsion between them is (A)  ( (1/3) )× 10-9N  (B)  ( (2/9) )× 10-9N  (C)  ( (1/9) )× 10-9N  (D)  ( (4/3) )× 10-9N

Tardigrade Admin · Accepted Answer

(Kq1/3)=10,(Kq2/1)=10 Kq1=30× 10-2 Kq2=10× 10-2 F=(Kq1q2/10-2) =((30× 10-2)(10× 10-2)/K× 10-2) F=(1/3)× 10-9N

Q. Two conducting spheres of radii $3 c m$ and $1 c m$ are separated by a distance of $10 c m$ in free space. If the spheres are charged to same potential of $10 V$ each, the force of repulsion between them is

$(\frac{1}{3}) \times 10^{- 9} N$

$(\frac{2}{9}) \times 10^{- 9} N$

$(\frac{1}{9}) \times 10^{- 9} N$

$(\frac{4}{3}) \times 10^{- 9} N$

$(\frac{2}{3}) \times 10^{- 9} N$

$\frac{K q _{1}}{3} = 10, \frac{K q _{2}}{1} = 10$
$K q_{1} = 30 \times 10^{- 2}$
$K q_{2} = 10 \times 10^{- 2}$
$F = \frac{K q _{1} q _{2}}{10 ^{- 2}}$
$= \frac{( 30 \times 10 ^{- 2} ) ( 10 \times 10 ^{- 2} )}{K \times 10 ^{- 2}}$
$F = \frac{1}{3} \times 10^{- 9} N$

Q. Two conducting spheres of radii 3cm and 1cm are separated by a distance of 10cm in free space. If the spheres are charged to same potential of 10V each, the force of repulsion between them is

(31​)×10−9N

(92​)×10−9N

(91​)×10−9N

(34​)×10−9N

(32​)×10−9N

3Kq1​​=10,1Kq2​​=10 Kq1​=30×10−2 Kq2​=10×10−2 F=10−2Kq1​q2​​ =K×10−2(30×10−2)(10×10−2)​ F=31​×10−9N