Question

A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net field $\vec{E}$ at the centre $O$ is

• A. $\frac{q}{4\,\pi^{2}\,\varepsilon_{0}\,r^{2}} \hat{j}$
• B. $-\frac{q}{4\,\pi^{2}\,\varepsilon_{0}\,r^{2}} \hat{j}$
• C. $-\frac{q}{2\,\pi^{2}\,\varepsilon_{0}\,r^{2}} \hat{j}$
• D. $\frac{q}{2\,\pi^{2}\,\varepsilon_{0}\,r^{2}} \hat{j}$

Answer 1

Answer: (C)

$-\frac{q}{2\,\pi^{2}\,\varepsilon_{0}\,r^{2}} \hat{j}$

Answer 2

Linear charge density $\lambda = \left(\frac{q}{\pi r}\right)$ $E = \int dE\,sin\,\theta \left(-\hat{j}\right) =\int \frac{K.dq}{r^{2}}sin\,\theta \left(-\hat{j}\right)$ $E = \frac{K}{r^{2}}\int\frac{qr}{\pi r} d\theta\,sin\,\theta \left(-\hat{j}\right)$ $= \frac{K}{r^{2}} \frac{q}{\pi } \int\limits^{\pi}_{0} \,sin\,\theta \left(-\hat{j}\right)$ $= \frac{q}{2\,\pi^{2}\,\varepsilon_{0}\,r^{2}} \left(-\hat{j}\right)$

Mathematically, the electrostatic force per unit charge exerted on a tiny unit positive charge at that place in space defines an electric field as a vector field. E=F/q is the formula for the electric field. Where

E = Intensity of the electric field generated by a point charge

test charge = q

F is the force that the test charge feels as a result of the point charge.

Think about two 'point' charges Q and q that are positioned in a system r distance apart. According to Coulomb's Law, which is provided by, the charge Q exerts a force on the charge q.

F = 1/4πɛ0 x Qq/r2

Now, E = F/q so, E = 1/4πɛ0 x Q/r2

Using the previous equation,

It follows that the Force F becomes equal to the Electric Field E produced by the charge Q itself if the magnitude of Q is taken to be unity, or 1.

Therefore, the force that a unit positive charge would encounter if it were put at that position may be represented as the Electric Field E caused by a charge Q at that point.

Here, Q is referred to as the source charge, and any additional charges q positioned nearby to assess the effects of the source are referred to as test charges.

The SI unit for electric field intensity is the Newton per Coulomb (NC-1).

The electric field's dimensional formula is [MLT-3A-1].