Question

At the centre of a half ring of radius $R = 10 \, \text{cm}$ and linear charge density $4n \, \text{C m}^{-1}$, the potential is $x \pi \, \text{V}$. The value of $x$ is ______.

Answer 1

Correct Answer: 36

To determine the electric potential at the center of a semicircular ring with radius \( R = 10 \, \text{cm} \) and linear charge density \( \lambda = 4 \, \text{nC} \, \text{m}^{-1} \), we apply the formula for the potential of a charged arc: \( V = \frac{k \lambda L}{R} \). Here, \( k \) is Coulomb's constant \( \left( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \right) \), \( L \) is the arc length of the semicircle, and \( R \) is the radius. For a semicircle, the arc length is \( L = \pi R \).

First, convert the linear charge density to coulombs per meter: \( \lambda = 4 \times 10^{-9} \, \text{C/m} \). The radius is \( R = 0.1 \, \text{m} \).

The arc length is calculated as \( L = \pi R = \pi \times 0.1 \, \text{m} = 0.1\pi \, \text{m} \).

Substitute these values into the potential formula:

\( V = \frac{(8.99 \times 10^9) \times (4 \times 10^{-9}) \times (0.1\pi)}{0.1} \).

Simplifying the expression yields \( V = 8.99 \times 4 \times \pi \).

The calculated potential is \( V = 35.96\pi \, \text{V} \).

We express this as \( V = x\pi \) to find \( x \), which gives \( x = 35.96 \).

Rounding \( x \) to the nearest whole number results in \( x = 36 \).

The computed value is 36, consistent with the expected range of 36,36.