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1. Electric Charges and Fields
normal
Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is
A
$\frac{q}{{2{\pi ^2}{\varepsilon _0}{R^2}}}$
B
$\frac{q}{{4{\pi ^2}{\varepsilon _0}{R^2}}}$
C
$\frac{q}{{4\pi {\varepsilon _0}{R^2}}}$
D
$\frac{q}{{2\pi {\varepsilon _0}{R^2}}}$
Solution
${\rm{E}} = \frac{{2{\rm{k}}\lambda }}{{\rm{r}}} = \frac{{2\lambda }}{{4\pi { \in _0}{\rm{r}}}} = \frac{{\rm{q}}}{{2{\pi ^2}{ \in _0}{{\rm{r}}^2}}}$ $ \quad\left[\because \lambda=\frac{\mathrm{q}}{\pi \mathrm{r}}\right]$
Standard 12
Physics
