A solid metallic sphere of radius r is charged to a voltage V. It is enclosed by a thin spherical shell of radius 2r. If q is the charge on the sphere and the shell each, the potential difference between them will be $\frac{x V}{2}$ . What is x?

___

Open in App

Solution

Charge on the inner sphere is, q = cV = $4 π ϵ_{\circ} r V$

The potential difference between the two spheres is,

V' = $V_{1} - V_{2}$ = $\frac{q}{4 π ϵ_{\circ}} (\frac{1}{r} - \frac{1}{2 r}) = \frac{q}{4 π ϵ_{\circ}} \frac{1}{2 r} = \frac{V}{2}$

Suggest Corrections

Similar questions

Q. If a spherical shell of radius r1, with charge q1, encloses a sphere of radius r2 with charge q2, what would be the electric potential of the outer shell due to the sphere it encloses?

Q. A metal sphere

A

of radius

a

is charged to potential

V

. What will be the its potential if it is enclosed by a spherical conducting shell

B

of radius

b

and the two are connected by metalic wire.

Q. A metal sphere A of radius

r_{1}

charged to a potential

ϕ_{1}

is enveloped by a thin walled conducting spherical shell B of radius

r_{2}

. Then potential

ϕ_{2}

of the sphere A after it is connected to the shell B by a thin conducting wire will be :-

Q. A metal sphere A of radius r₁charged to a potential phi₁is enveloped by a thin walled conducting spherical shell B of radius r₂. Then potential phi₂of the sphere A after it is connected to the shell B by a thin conducting wire will be :-

A metallic sphere of radius $R_{1}$ is charged to a potential V. If it is enclosed by a spherical conducting shell of radius $R_{2}$ and connected to it, its new potential will be

Join BYJU'S Learning Program

A solid metallic sphere of radius r is charged to a voltage V. It is enclosed by a thin spherical shell of radius 2r. If q is the charge on the sphere and the shell each, the potential difference between them will be x V2. What is x? ___

Charge on the inner sphere is, q = cV = 4πϵ∘rV The potential difference between the two spheres is, V' = V1−V2 = q4πϵ∘(1r−12r)=q4πϵ∘12r=V2

A solid metallic sphere of radius r is charged to a voltage V. It is enclosed by a thin spherical shell of radius 2r. If q is the charge on the sphere and the shell each, the potential difference between them will be $\frac{x V}{2}$ . What is x?

___

Charge on the inner sphere is, q = cV = $4 π ϵ_{\circ} r V$

The potential difference between the two spheres is,

V' = $V_{1} - V_{2}$ = $\frac{q}{4 π ϵ_{\circ}} (\frac{1}{r} - \frac{1}{2 r}) = \frac{q}{4 π ϵ_{\circ}} \frac{1}{2 r} = \frac{V}{2}$