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चार बंद पृष्ठ तथा उनके आवेश विन्यास को निम्न चित्र में दर्शाया गया है।
यदि उनके पृष्ठ से बद्ध वैद्युत फ्लक्स क्रमशः $\Phi_{1}, \Phi_{2^{\prime}} \Phi_{3}$ तथा $\Phi_{4}$ हों तो
${\phi _1} < {\phi _2} = {\phi _3} > {\phi _4}$
${\phi _1} > {\phi _2} > {\phi _3} > {\phi _4}$
${\phi _1} = {\phi _2} = {\phi _3} = {\phi _4}$
${\phi _1} > {\phi _3} ; {\phi _2} < {\phi _4}$
Solution
The net fluw linked with closed surfaces $S_{1}$, $\mathrm{S}_{2}, \mathrm{S}_{3} \& \mathrm{S}_{4}$ are
For surface $S_{1}, \phi_{1}=\frac{1}{\varepsilon_{0}}(2 q)$
For surface
$\mathrm{S}_{2}, \phi_{2}=\frac{1}{\varepsilon_{0}}(\mathrm{q}+\mathrm{q}+\mathrm{q}-\mathrm{q})=\frac{1}{\varepsilon_{0}} 2 \mathrm{q}$
For surface $S_{3}, \phi_{3}=\frac{1}{\varepsilon_{0}}(q+q)=\frac{1}{\varepsilon_{0}}(2 q)$
For surface
$S_{4}, \phi_{4}=\frac{1}{\varepsilon_{0}}(8 q-2 q-4 q)=\frac{1}{\varepsilon_{0}}(2 q)$
Hence, $\phi_{1}=\phi_{2}=\phi_{3}=\phi_{4}$ i.e. netelectric flux is same for all surfaces.
Keep in mind, the electric field due to a charge outside $\left(\mathrm{S}_{3} \text { and } \mathrm{S}_{4}\right),$ the Gaussian surface contributes zero net flux through the surface, because as many lines due to that charge enter the surface as leave it.
