Question

A small sphere of mass m and carrying a charge q attached to one end of an insulating thread of leng a, the other end of which is fixed at 0.0 as shown in figure. There exists a uniform electric field $\vec{E} = -E_0\hat{j}$ in the region. The minimum velocity which should be given to the sphere at 4.0 in the direction shown so that it is able to complete the circle around the origin is (There is no gravity)

• A. $\sqrt{\frac{5qE_0a}{m}}$
• B. $\sqrt{\frac{3qE_0a}{m}}$
• C. $\sqrt{\frac{qE_0a}{m}}$
• D. $2\sqrt{\frac{qE_0a}{m}}$

Answer 1

Answer: (B)

$\sqrt{\frac{3qE_0a}{m}}$

Answer 2

We analyze the motion as a constrained circular motion under a conservative “electric weight” force.

Let the sphere (mass m, charge q) be on a circle of radius a. Its potential energy in the uniform field ( 𝐸 = –E₀ 𝑗 ) is given by

  U = qE₀ y = qE₀ a sinθ

(with the pivot O at the center and at (0,0); note that y = a sinθ).

A velocity v₀ (to be determined) is imparted at θ = 0 (point (a,0)) in the upward direction (tangent to the circle). By energy conservation, the energy at any angle θ is

  ½ m v²(θ) + qE₀ a sinθ = ½ m v₀².   (1)

For complete circular motion the tension must never vanish. Writing the radial force balance we have   T + (component of electric force along the radius) = m v²(θ)/a. The radial unit vector (pointing from O to the sphere) is (cosθ, sinθ). The electric force is 𝐹ₑ = –qE₀ 𝑗 so its radial component is

  F_r = 𝐹ₑ · (cosθ, sinθ) = –qE₀ sinθ.

Thus   T = m v²(θ)/a + qE₀ sinθ. For the string to remain taut, T ≥ 0 at all θ. The most “dangerous” case is when sinθ is maximum (i.e. sinθ = 1 at θ = π/2):   T(π/2) = m v²(π/2)/a + qE₀ ≥ 0  ⇒ m v²(π/2)/a ≥ qE₀.  (2)

From (1) at θ = π/2:   ½ m v²(π/2) + qE₀ a = ½ m v₀²  ⇒  v²(π/2) = v₀² – (2qE₀ a/m).  (3)

Substitute (3) into (2):   m (v₀² – (2qE₀ a/m))/a ≥ qE₀  ⇒   (m v₀²)/a – 2qE₀ ≥ qE₀  ⇒   (m v₀²)/a ≥ 3qE₀  ⇒   v₀² ≥ (3qE₀ a)/m.

Thus the minimum speed is   v₀(min) = √(3qE₀a/m).