- Home
- Standard 12
- Physics
2. Electric Potential and Capacitance
medium
A hemispherical bowl of mass $m$ is uniformly charged with charge density $'\sigma '$ . Electric potential due to charge distribution at a point $'A'$ is (which lies at centre of radius as shown).
A
$\frac{{\sigma R}}{{4{\varepsilon _0}}}$
B
$\frac{{\sigma R}}{{3{\varepsilon _0}}}$
C
$\frac{{\sigma R}}{{2{\varepsilon _0}}}$
D
$\frac{{\sigma R}}{{{\varepsilon _0}}}$
Solution
$V = \frac{{{\rm{kQ}}}}{{\rm{R}}} = \frac{{\sigma \times 2\pi {{\rm{R}}^2}}}{{4\pi { \in _0}{\rm{R}}}} = \frac{{\sigma {\rm{R}}}}{{2{ \in _0}}}$
Standard 12
Physics
Similar Questions
medium
