Q.
A sphere of 4cm radius is suspended within hollow sphere of 6cm radius. The inner sphere is charged to a potential 3esu, when the outer sphere is earthed. The charge on the inner sphere is :
The sphere of radius =4cm=4×10−2m
hollow sphere =6cm=6×10−2m
inner sphere charged to a potential 3 e.s.u.
lesu =300V
lesu =3×10−10C
Let, Charge on inner sphere is +Q, then, −Q is induced in the surface of outer sphere.
Now, potential on inner sphere =r1Q1+r2Q2
Here, Q1 is the charge on inner sphere. r1 is the radius of inner sphere. Q2 is the charge on outer sphere. r2= the radius of the outer sphere.
Given, Potential on inner sphere =3 esu of potential r1=4cm r2=6cm 3 esu =4Q−6Q ⇒Q[41−61]=3
or, 12Q=3
or, Q=36 esu
Hence, charge =36 esu of charge.