A hollow metallic sphere of radius $R$ is given a charge $Q$ . Then, the potential at the centre is :

zero

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{4 π ε_{0}} . \frac{Q}{R}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{4 π ε_{0}} . \frac{2 Q}{R}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{4 π ε_{0}} . \frac{Q}{2 R}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{1}{4 π ε_{0}} . \frac{Q}{R}$
Potential at the center of hollow metallic sphere is same as the potential at its surface i.e. $\frac{1}{4 π ϵ_{o}} . \frac{Q}{R}$

Suggest Corrections

Similar questions

Q. A hollow metallic sphere of radius R is given a charge Q. The potential at its centre is

Q. A system comprises of a solid non conducting sphere of charge

Q

and radius

R

. A point charge

q

is placed at a distance

r

from the centre of sphere. Find the total energy of the system

[k = \frac{1}{4 π ε_{0}}]

A point charge q is placed inside a conducting spherical shell of inner radius 2R and outer radius 3R at a distance of R from the centre of the shell. The electric potential at the centre of shell will be times.

A ring with a uniform charge $Q$ and radius $R$ , is placed in the $y - z$ plane with its centre at the origin $(k = \frac{1}{4 π ε_{0}})$ . Then :

A half ring of radius $R$ has a charge of $λ$ per unit length. The potential at the centre of the half ring is $(k = \frac{1}{4 π ε_{0}})$

Join BYJU'S Learning Program

A hollow metallic sphere of radius R is given a charge Q. Then, the potential at the centre is :

The correct option is B 14πε0.QRPotential at the center of hollow metallic sphere is same as the potential at its surface i.e. 14πϵo.QR

A hollow metallic sphere of radius $R$ is given a charge $Q$ . Then, the potential at the centre is :

The correct option is B $\frac{1}{4 π ε_{0}} . \frac{Q}{R}$
Potential at the center of hollow metallic sphere is same as the potential at its surface i.e. $\frac{1}{4 π ϵ_{o}} . \frac{Q}{R}$