Question

A small sphere of mass m and carrying a charge q attached to one end of an insulating thread of length 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: (A)

$\sqrt{\frac{5qE_0a}{m}}$

Answer 2

Solution:

1. Identify Effective “Gravity”:
   The electric field is given by

   $$\vec{E} = -E_0 \hat{j}.$$

   A particle with charge $q$ experiences a force

   $$\vec{F} = q\vec{E} = -qE_0 \hat{j}.$$

   This force acts like gravity with effective acceleration

   $$g_{\rm eff} = \frac{qE_0}{m}.$$

2. Energy Considerations for Circular Motion:
   In a pendulum (or circular motion) problem, if you give an initial speed at the lowest point, to complete the circle the speed at the highest point must be at least such that the tension is zero. At the top, the minimal condition is that the required centripetal force is provided solely by the effective weight:

   $$\frac{mv_{\text{top}}^2}{a} = m g_{\rm eff} \quad \Longrightarrow \quad v_{\text{top}}^2 = g_{\rm eff}a = \frac{qE_0a}{m}.$$

3. Energy Conservation from Bottom to Top:
   Let the kinetic energy at the bottom be

   $$KE_{\rm bottom} = \frac{1}{2} m v_{\rm min}^2,$$

   and at the top, the sphere gains potential energy. For a circular path of radius $a$, the vertical height difference between the bottom and the top is $2a$ so that

   $$PE_{\rm top} = m g_{\rm eff} (2a) = 2a \frac{qE_0}{m} m = 2qE_0 a.$$

   Also, the kinetic energy at the top must be at least

   $$KE_{\rm top} = \frac{1}{2} m v_{\text{top}}^2 = \frac{1}{2} m \left(\frac{qE_0a}{m}\right) = \frac{qE_0a}{2}.$$

   So by conservation of energy:

   $$\frac{1}{2}m v_{\rm min}^2 = 2qE_0a + \frac{qE_0a}{2} = \frac{5qE_0a}{2}.$$

4. Solving for $v_{\rm min}$:

   $$v_{\rm min}^2 = \frac{5qE_0a}{m} \quad \Longrightarrow \quad v_{\rm min} = \sqrt{\frac{5qE_0a}{m}}.$$

Thus, the minimum velocity is

$$\sqrt{\frac{5qE_0a}{m}}.$$

Brief Explanation:

• Replace gravity with the effective acceleration $g_{\rm eff} = \frac{qE_0}{m}$.
• At the top, minimal speed is $\sqrt{g_{\rm eff}a}$.
• Use energy conservation: difference in potential energy $=2qE_0a$ plus kinetic energy at top $\frac{qE_0a}{2}$ equals kinetic energy at the bottom.
• Solve to get $v_{\rm min} = \sqrt{\frac{5qE_0a}{m}}$.