Practicing Success
Spherical symmetric charge system centered at origin. Electric potential $\phi=\frac{Q}{4 \pi \varepsilon_0 R_0} \quad r \leq R_0$ $\phi=\frac{Q}{4 \pi \varepsilon_0 r} \quad r>R_0$ |
A spherical symmetry at r = 2R0 encloses a net charge Q Electric field is discontinued at r = R0 Change is only present at r = R0 All of the above |
All of the above |
For $r>R_0, E=-\frac{d \phi}{d r}=\frac{Q}{4 \pi \varepsilon_0 r^2}$ Charge enclosed by the concentric spherical surface of $r=2 R_0$ $=\varepsilon_0 \phi_{C}=\varepsilon_0 E \times 4 \pi \times r^2=\varepsilon_0 \frac{Q}{4 \pi \varepsilon_0 r^2} 4 \pi r^2=Q$ For $r<R_0, E=-\frac{d \phi}{dr}=0$ (∵ c = constant) For $r>R_0, E=-\frac{d \phi}{d t}=\frac{Q}{4 \pi \varepsilon_0 r^2}$ Since for $r<R_0, E=0$, hence charge will be only on the spherical surface of $r=R_0$. For $r<R_0, E=0 \Rightarrow$ energy density $\frac{1}{2} \varepsilon_0 E^2=0$ (d) |