A thin wire ring of radius R has an electric charge q. What will be the increment of the force stretching the wire…

Problem1:  A thin wire ring of radius R has an electric charge q. What will be the increment of the force stretching the wire if a point charge q0 is placed at the centre of the ring ?

Solution :

As we put a charge q0 at the centre of the ring, the wire get stretched due to mutual repulsion between the positive charge q0 and each element of the wire having positive charge (dq). Such a small element is shown in the diagram.

Tensions acting at the ends of the elements due to elasticity of the ring are shown. Its two components are also shown. Carefully note the geometry.

dF is the repulsive force between the elemental charge dq and qo.

Hence, the radial forces acting on the element can be summed up as;

$ \displaystyle dF – 2T sin(\frac{d\theta}{2}) = 0 $

$ \displaystyle dF = 2T sin(\frac{d\theta}{2}) $

$ \displaystyle dF = 2T (\frac{d\theta}{2}) $ (for small angle , sin(dθ/2) = dθ/2 )

dF = Tdθ

Now evidently

$ \displaystyle dF = \frac{q_0 dq}{4\pi \epsilon_0 R^2} $ and $ \displaystyle dq = \frac{q}{2\pi R}(R d\theta) $

Hence , $ \displaystyle dF = \frac{q_0 q d\theta}{8 \pi^2 \epsilon_0 R^2} $

$\displaystyle T d\theta = \frac{q_0 q d\theta}{8 \pi^2 \epsilon_0 R^2} $

$\displaystyle T = \frac{q_0 q}{8 \pi^2 \epsilon_0 R^2} $

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