English: So, I have written all terms, this is equal to V d minus x over d, x will now become x naught, because of this delta function times Q plus V Q 1 and this indefinitely gives you Q 1 equals minus Q, d minus $0\ = \frac{V(dx)}{d}Q\delta\left( \overrightarrow{r} - x\widehat{x} \right)dV\ + \ Vq_{1} + 0$ So, if I look at this capacitor again, this distance was x 0 and the charge density or Q 1 on this is equal to nothing but Q the distance from the other plate divide by d, the minus sign.
Telugu: V₁ρ₂dV + V₁^(surface)σ₁dS = V₂ρ₁dV + V₂^(surface)σ₁dS $q_{1} = \ - Q\frac{(d - x_{0})}{d}$ కాబట్టి, ఒకవేళ నేను మళ్ళీ ఈ కెపాసిటర్‌ను చూస్తే, ఈ దూరం x 0 గా ఉంది మరియు దీనిపై ఆవేశ సాంద్రత లేదా Q 1 అనునది మరేమీ.