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A small charged pith ball of mass m and charge q is suspended by a light insulating string. A uniform horizontal electric field of magnitude E is switched on. In equilibrium, the angle the string makes with the vertical is :

Options

  1. A⁻¹ ( qE mg )
  2. B⁻¹ ( mg qE )
  3. C⁻¹ ( qE mg )
  4. D⁻¹ ( qE mg )

Correct answer

A. ⁻¹ ( qE mg )

Step-by-step solution

Let T be the tension in the string and be the angle it makes with the vertical in equilibrium. The forces acting on the pith ball are: 1. Weight mg acting vertically downwards. 2. Electric force qE acting horizontally. 3. Tension T acting along the string. For the ball to be in equilibrium, the net force in both horizontal and vertical directions must be zero. We resolve the tension T into its components: Vertical component: T = mg Horizontal component: T = qE Dividing the horizontal equilibrium equation by the ver

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