Define electric flux. Write its SI …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Maahnoor Quraishi 2 years, 1 month ago
- 2 answers
Mayank Chauhan 2 years, 1 month ago
Related Questions
Posted by Ashfak Altaf Khan 7 months ago
- 1 answers
Posted by Aniket Mahajan 10 hours ago
- 0 answers
Posted by Priyanka Goswami 7 months ago
- 0 answers
Posted by Khushi Keshyap Kuhu 7 months ago
- 0 answers
Posted by Khushbu Otti 7 months, 1 week ago
- 0 answers
Posted by Savitri A 1 week, 6 days ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Preeti Dabral 2 years, 1 month ago
The charge resides on the surface of a conductor. Thus, a hollow charged conductor is equivalent to a charged spherical shell. Let's consider a spherical Gaussian surface of radius (r < R). If E is the electric field inside the shell, then by symmetry electric field strength has the same magnitude Ei on the Gaussian surface and is directed radially outward.
Electric flux through the Gaussian surface is given by,
{tex}=\int_{s} \vec{E}_{i} \cdot d \vec{S}{/tex}
{tex}=\int E_{i} d S \cos 0=E_{i} .4 \pi r^{2}{/tex}
Now, Gaussian surface is inside the given charged shell, so charge enclosed by Gaussian surface is zero.
Therefore, using Gauss's theorem, we have
{tex}\int_{S} \vec{E}_{i} \cdot d \vec{S}=\frac{1}{\epsilon_{0}} \times \text { charge enclosed }{/tex}
{tex}\Rightarrow E_{i} \cdot 4 \pi r^{2}=\frac{1}{\epsilon_{0}} \times 0{/tex}
{tex}\Rightarrow{/tex} Ei = 0
Thus, electric field at each point inside a charged thin spherical shell is zero.
0Thank You