Q વિદ્યુતભાર ધરાવતા એક ધન વાહક ગોળોએ અવિદ્યુત | vedclass

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Question

The electric field in between the shell and sphere is

$E . \pi x ^2=\frac{ Q _{ en }}{\epsilon_0}=\frac{ Q }{\epsilon_0}$ (using Gauss's law)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

The electric field in between the shell and sphere is

$E . \pi x ^2=\frac{ Q _{ en }}{\epsilon_0}=\frac{ Q }{\epsilon_0}$ (using Gauss's law)

$E=\frac{Q}{4 \pi \epsilon_0 x^2}$

The potential difference between the shells is $d V=V_r-V_R=\int \limits_r^R E d x=$

$\int \limits_r^R \frac{Q}{4 \pi \epsilon_0 x^2} d x=\frac{Q}{4 \pi \epsilon_0}(1 / r-1 / R)$

Thus, $V =\frac{ Q }{4 \pi \varepsilon_0}(1 / r -1 / R )$

As the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere, Since the two values have not changed, potential difference does not change. Hence the potential difference remains $V$.

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