Question

A pendulum bob of mass 80 mg and carrying a charge of 2X10 -8 C is at rest in a uniform, horizontal electric field of 20 kVm -1 . Find the tension in the thread.

Answer 1

see diagram.
  
    From the equilibrium of concurrent forces, 
 
       T / Sin 90   =   q * E / Sin (180-α) = m g / Sin(90+α)
   or,    T = q * E / Sin α = m g / Cos α
 
     tan α = q E / mg
              = 2 * 10⁻⁸ * 20 * 10³ / [80 * 10⁻³ * 9.81 ] = 0.000509
        α = 0.292 deg.
      T = mg / cosα = 0.080 * 9.81 / 0.999987 = 0.7846 Newtons

Answer 2

Answer:8.8×10^-4

Explanation:Mass =80mg

Charge =2×10^−8 Coulomb

Electric field =20,000Vm^−1

Given, mass =80mg=8×10^−6kg

charge =q=2×10^−8C

Let T be the tension in string and θ is the angle if string with vertical.

In equilibrium

Tsinθ=qE⟶(1)

Tcosθ=mg⟶(2)

Divide (1) by (2) we get

tanθ= mgqE

⇒tanθ= 80×10^−6×10^2×10^−8×2×10^4

tanθ=0.5

θ=tan^−1(0.5)

θ=27°

Now,

tension, T= sinθ

qE =sin27°

2×10^−8×2×10^4

=8.8×10^−4V