Pendulum Equilibrium Force Analysis

When a charged pendulum bob is placed in a uniform horizontal electric field, it experiences three forces at equilibrium:

• Gravitational force: $ \vec{F}_g = m\vec{g} $ (acting vertically downwards)
• Electric force: $ \vec{F}_e = q\vec{E} $ (acting horizontally)
• Tension force: $ \vec{T} $ (acting along the string)

Tension Calculation at Equilibrium

At equilibrium, the pendulum string makes an angle $ \theta $ with the vertical. We resolve the tension $ \vec{T} $ into its vertical and horizontal components:

• Vertical component: $ T_v = T \cos(\theta) $
• Horizontal component: $ T_h = T \sin(\theta) $

For the bob to be in equilibrium, the net force must be zero. Thus, the vertical and horizontal components must balance the respective applied forces:

1. Vertical balance: $ T \cos(\theta) = mg $
2. Horizontal balance: $ T \sin(\theta) = qE $

Deriving Final Tension

To find the magnitude of the tension $ T $, we square both equations and add them:

$ (T \cos(\theta))^2 + (T \sin(\theta))^2 = (mg)^2 + (qE)^2 $

$ T^2 \cos^2(\theta) + T^2 \sin^2(\theta) = m^2g^2 + q^2E^2 $

Factor out $ T^2 $:

$ T^2 (\cos^2(\theta) + \sin^2(\theta)) = m^2g^2 + q^2E^2 $

Using the trigonometric identity $ \cos^2(\theta) + \sin^2(\theta) = 1 $:

$ T^2 = m^2g^2 + q^2E^2 $

Taking the square root to find the tension $ T $:

$ T = \sqrt{m^2g^2 + q^2E^2} $

This corresponds to the magnitude of the tension in the string when the pendulum attains equilibrium.