Consider two conducting spheres of radii $R_{1}$ and $R_{2}$ with $R_{1} > R_{2}$ . If the two are at the same potential, and the larger sphere has more charge than the smaller sphere, then

the charge density of smaller sphere is less than that of larger sphere.

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

the charge density of smaller sphere is more than that of larger sphere.

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

both spheres may have same charge density.

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

None of these.

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

Open in App

Solution

The correct option is D the charge density of smaller sphere is more than that of larger sphere.
Here, $V_{1} = V_{2} o r \frac{q_{1}}{4 π ε_{0} R_{1}} = \frac{q_{2}}{4 π ε_{0} R_{2}}$
$∴ \frac{q_{1}}{q_{2}} = \frac{R_{1}}{R_{2}} . . . (i)$

Given $R_{1} > R_{2}$

$∴ q_{1} > q_{2}$

$∴$ Larger sphere has more charge than the smaller sphere. Now charge densities
$σ_{1} = \frac{q_{1}}{4 π R_{1}^{2}}; σ_{2} = \frac{q_{2}}{4 π R_{2}^{2}}$

$∴ \frac{σ_{2}}{σ_{1}} = \frac{q_{2}}{q_{1}} \frac{R_{2}^{2}}{R_{2}^{2}} o r \frac{σ_{2}}{σ_{1}} = \frac{R_{2}}{R_{1}} \frac{R_{1}^{2}}{R_{2}^{2}} = \frac{R_{1}}{R_{2}}$ (using (i))

As $R_{1} > R_{2}$ , therefore $σ_{2} > σ_{2}$
Charge density of smaller sphere is more than the charge density of larger sphere

Suggest Corrections

Similar questions

Two metal spheres of radii $R_{1}$ and $R_{2}$ are charged to the same potential. The ratio of the charge on the two spheres is:

r_{1}

r_{2}

r_{1}

r_{2}

q_{1}

q_{2}

(σ)

Two conducting spheres of radii $r_{1}$ and r₂ are at the same potential. Their charges are in the ratio of

R_{1}

R_{2}

Join BYJU'S Learning Program