From the image, consider a point $P$ inside the cavity and let $\rho$ be the charge density. Then, after applying superposition principle the net electric field at point $P$ will be:
$E=\frac{\rho}{3 \epsilon_0} \vec{b}-\frac{\rho}{3 \epsilon_0} \vec{a}$
$=\frac{\rho}{3 \epsilon_0} \vec{r} \text { (Since } \vec{r}=\vec{b}-\vec{a} \text { ) }$
which for a given cavity is independent of the position of a point within the cavity.
Therefore, Electric field is non-zero and uniform within the cavity.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$T_{1}=27^{\circ} C$ [outside fridge]
$T_{2}=-23^{\circ} C$ [inside fridge]
Considering only $P-V$ work is involved, the total change in enthalpy (in Joule) for the transformation of state in the sequence $X \rightarrow Y \rightarrow Z$ is $\qquad$
[Use the given data: Molar heat capacity of the gas for the given temperature range, $C _{ v , m }=12 J K ^{-1} mol ^{-1}$ and gas constant, $R =8.3 J K ^{-1} mol ^{-1}$ ]
