Consider two charged metallic spheres $1$ and $2$ of radii $R_{1}$ and $R_{2}$ , respectively. The electric fields $E_{1}$ (on $S_{1}$ ) and $E_{2}$ (on $S_{2}$ ) on their surfaces are such that $\frac{E_{1}}{E_{2}} = \frac{R_{1}}{R_{2}}$ . Then the ratio $\frac{V_{1} (o n S_{1})}{V_{2} (o n S_{2})}$ of the electrostatic potentials on each sphere is

$\frac{R_{1}}{R_{2}}$

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

${(\frac{R_{1}}{R_{2}})}^{2}$

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

$\frac{R_{2}}{R_{1}}$

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

${(\frac{R_{1}}{R_{2}})}^{3}$

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

Open in App

Solution

The correct option is B
${(\frac{R_{1}}{R_{2}})}^{2}$

Step 1. Given data
$\frac{E_{1}}{E_{2}} = \frac{R_{1}}{R_{2}}$
Step 2. Applying the formula of the Electrostatic potential, $V$
$V = E \times R$ [Where, $E$ is the electric field and $R$ is the radius.]
Ratio of the Electrostatic potential of both the spheres
$\frac{V_{1}}{V_{2}} = \frac{E_{1}}{E_{2}} \times \frac{R_{1}}{R_{2}}$
$\frac{V_{1}}{V_{2}} = \frac{R_{1}}{R_{2}} \times \frac{R_{1}}{R_{2}}$
$\frac{V_{1}}{V_{2}} = {(\frac{R_{1}}{R_{2}})}^{2}$
Hence, the correct option is $(B)$ .

Suggest Corrections

Similar questions

r_{1}

r_{2}

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}

Join BYJU'S Learning Program