Assertion : Two concentric charged shells are | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{1}+Q_{2}}{R_{2}}$

$\mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_{0}}\left(\

Vedclass · Accepted Answer

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Vedclass · Answer

$\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{1}+Q_{2}}{R_{2}}$

$\mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{Q_{1}}{R_{1}}+\frac{Q_{2}}{R_{2}}\right)$

$\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{0}} Q_{1}\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

Assertion : Two concentric charged shells are given. The potential difference between the shells depends on charge of inner shell.

Reason : Potential due to charge of outer shell remains same at every point inside the sphere.

Similar Questions

$N$ identical spherical drops charged to the same potential $V$ are combined to form a big drop. The potential of the new drop will be

The radius of a soap bubble whose potential is $16\,V$ is doubled. The new potential of the bubble will be.....$V$

The give graph shown variation (with distance $r$ from centre) of

A hollow conducting sphere of radius $R$ has a charge $( + Q)$ on its surface. What is the electric potential within the sphere at a distance $r = \frac{R}{3}$ from its centre