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Question

(B) The energy density $U$ in an electric field is given by the formula $U = \frac{1}{2} \varepsilon_0 E^2$,where $E$ is the electric field intensity.

Vedclass · Accepted Answer

$U_A > U_B > U_C$

Vedclass · Answer

(B) The energy density $U$ in an electric field is given by the formula $U = \frac{1}{2} \varepsilon_0 E^2$,where $E$ is the electric field intensity.For an isolated conducting sphere of radius $r$ carrying charge $Q$,the electric field just outside its surface is $E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}$.Substituting this into the energy density formula: $U = \frac{1}{2} \varepsilon_0 \left( \frac{Q}{4 \pi \varepsilon_0 r^2} \right)^2 = \frac{Q^2}{32 \pi^2 \varepsilon_0 r^4}$.Since $Q$ is constant,we have $U \propto \frac{1}{r^4}$.Given the radii $R_A = R$,$R_B = 2R$,and $R_C = 3R$,we find:$U_A \propto \frac{1}{R^4}$,$U_B \propto \frac{1}{(2R)^4} = \frac{1}{16R^4}$,and $U_C \propto \frac{1}{(3R)^4} = \frac{1}{81R^4}$.Comparing these values,it is clear that $U_A > U_B > U_C$.

Three isolated metal spheres $A$,$B$,and $C$ have radii $R$,$2R$,and $3R$ respectively,and the same charge $Q$. If $U_A$,$U_B$,and $U_C$ are the energy densities just outside the surfaces of the spheres,then the relation between $U_A$,$U_B$,and $U_C$ is:

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