MCQ
A semicircular ring of radius $'a'$ has charge density $\lambda  = {\lambda _0}\,\cos \,\theta $ where ${\lambda _0}$ is constant and $'\theta'$ is shown in figure. Then total charge on the ring is
  • A
    $2{\lambda _0}\,R$
  • B
    $4{\lambda _0}\,R$
  • C
    $\frac{{{\lambda _0}R}}{3}$
  • Zero

Answer

Correct option: D.
Zero
d
$\mathrm{q}=\int_{0}^{\pi} \lambda \mathrm{Rd} \theta \quad \lambda=\lambda_{0} \cos \theta$

$\mathrm{q}=\lambda_{0} \mathrm{R} \int_{0}^{\pi} \cos \theta \mathrm{d} \theta$

$\mathrm{q}=\lambda_{0} \mathrm{R}(\sin \theta)_{0}^{\pi}=0$

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