A disc of mass \( m = 2\text{ kg}\) and Radius \( R = 2\text{ m}\) is kept at the edge of the table. A non-uniform charge is distributed on the circumference of the disc whose linear charge density varies as \( \large{\lambda = \dfrac{\sin\theta}{\pi}} \)( \(\theta\) is measured from point \(A\)). A uniform horizontal electric field exist in the region as shown.

What is the angle the disc makes with the vertical when it loses contact with the table.

Your answer can be represented as \({ \phi = {\cos}^{-1}(\dfrac{a}{b\sqrt {c}}) - \dfrac{\pi}{4}} \).

Enter your answer as \( a \times b \times c \).

**Details and Assumptions** :

Electric field \( E = 17.5 N/C\), take \( g = 10 m/s^2\)

Angle ( \( \phi\) ) is measured clockwise.

Disc is not a conductor.

a , b and c are positive integers and c is a square free integer.

Friction is sufficient.

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