Electrostatics Question 5
Question 5 - 2024 (27 Jan Shift 1)
A thin metallic wire having cross sectional area of $10^{-4} m^{2}$ is used to make a ring of radius $30 cm$. A positive charge of $2 \pi C$ is uniformly distributed over the ring, while another positive charge of $30 pC$ is kept at the centre of the ring. The tension in the ring is $N$; provided that the ring does not get deformed (neglect the influence of gravity). (given, $\frac{1}{4 \pi \epsilon _0}=9 \times 10^{9} SI$ units)
Show Answer
Answer: (3)
$2 T \sin \frac{d \theta}{2}=\frac{kq _0}{R^{2}} \cdot \lambda Rd \theta$
$\left[\lambda=\frac{Q}{2 \pi R}\right]$
$\Rightarrow T=\frac{Kq _0 Q}{\left(R^{2}\right) \times 2 \pi}$
$=\frac{\left(9 \times 10^{9}\right)\left(2 \pi \times 30 \times 10^{-12}\right)}{(0.30)^{2} \times 2 \pi}$
$=\frac{9 \times 10^{-3} \times 30}{9 \times 10^{-2}}=3 N$
