Question

solution plx.......... ​

Answer 1

Answer:

I hope my answer helps u...

Answer 2

Answer:

2:1

At equilibrium, potential across both the spheres is same because they are connected by a wire

Let r' : 1 x $10^{-3}$

      r : 2 x $10^{-3}$

    Q' : charge on small sphere

     Q : charge on bigger sphere

So,

$\frac{KQ}{r}+\frac{KQ'}{R} = \frac{KQ'}{r'}+\frac{KQ}{R}$

$\frac{Q}{r}+\frac{Q'}{R} = \frac{Q'}{r'}+\frac{Q}{R}$

$\frac{Q}{2X10^{-3}}+\frac{Q'}{5X10^{-2}} = \frac{Q'}{1X 10^{-3}}+\frac{Q}{5X10^{-2}}$

On solving,

$\boxed{\frac{Q'}{Q}=\frac{24}{49}}$

Electric field on a sphere is due to:

1. Charge on itself
2. Charge on the other sphere

So electric field on $1^{st}$ sphere : $\frac{KQ'}{r'^{2}} -\frac{KQ}{R^{2}}$

So electric field on $2^{nd}$ sphere : $\frac{KQ}{r^{2}} -\frac{KQ'}{R^{2}}$

Ratio of electric field :

$\frac{\frac{KQ'}{r'^{2}} -\frac{KQ}{R^{2}}}{\frac{KQ'}{r'^{2}} -\frac{KQ}{R^{2}}}$

$\frac{\frac{Q'}{r'^{2}} -\frac{Q}{R^{2}}}{\frac{Q}{r^{2}} -\frac{Q'}{R^{2}}}$

On solving,

$\frac{2500Q'-Q}{625Q-Q'}$

$\frac{2500(\frac{24}{49} )Q-Q}{625Q-(\frac{24}{49} )Q}\\\\\frac{2500(\frac{24}{49} )-1}{625-(\frac{24}{49} )}\\\\\frac{59,951}{30,601}\\\\\boxed{\boxed{2:1}}$

Hope this answer helped you :)

[P.S : you could ignore the electric field due to other sphere ∵ of the large distance between two spheres as compared to their small radius]