Question

A small sphere carrying a charge 'q' is hanging in between two parallel plates by a string of length L. Time period of pendulum is To. When parallel plates are charged, the time period changes to T. The ratio T/To is equal to

• A. $\left(\frac{g}{g+\frac{qE}{m}}\right)^{1/2}$
• B. $\left(\frac{g}{g+\frac{qE}{m}}\right)^{3/2}$
• C. $\left(\frac{g}{g+\frac{qE}{m}}\right)^{1/2}$
• D. None of these

Answer 1

Answer: (A), (C)

$\left(\frac{g}{g+\frac{qE}{m}}\right)^{1/2}$

Answer 2

The time period of a simple pendulum of length L in a gravitational field with acceleration $g$ is given by $T_0 = 2\pi \sqrt{\frac{L}{g}}$.

When the pendulum is placed between charged parallel plates, an additional electric force acts on the charged sphere. Let's assume the electric field between the plates is uniform and vertical, with magnitude E. Let the charge on the sphere be q.

Case 1: Electric field is vertically downwards. The electric force on the sphere is $F_E = qE$ downwards (assuming q is positive). The gravitational force is $F_g = mg$ downwards. The total downward force is $F_{net} = F_g + F_E = mg + qE$. The effective acceleration is $g_{eff} = \frac{F_{net}}{m} = \frac{mg + qE}{m} = g + \frac{qE}{m}$. The time period in the presence of the electric field is $T = 2\pi \sqrt{\frac{L}{g_{eff}}} = 2\pi \sqrt{\frac{L}{g + \frac{qE}{m}}}$. The ratio $T/T_0 = \frac{2\pi \sqrt{\frac{L}{g + \frac{qE}{m}}}}{2\pi \sqrt{\frac{L}{g}}} = \sqrt{\frac{L/(g + qE/m)}{L/g}} = \sqrt{\frac{g}{g + \frac{qE}{m}}} = \left(\frac{g}{g + \frac{qE}{m}}\right)^{1/2}$.

Case 2: Electric field is vertically upwards. The electric force on the sphere is $F_E = qE$ upwards (assuming q is positive). The gravitational force is $F_g = mg$ downwards. The net downward force is $F_{net} = mg - qE$ (assuming $mg > qE$). The effective acceleration is $g_{eff} = \frac{F_{net}}{m} = \frac{mg - qE}{m} = g - \frac{qE}{m}$. The time period in the presence of the electric field is $T = 2\pi \sqrt{\frac{L}{g_{eff}}} = 2\pi \sqrt{\frac{L}{g - \frac{qE}{m}}}$. The ratio $T/T_0 = \frac{2\pi \sqrt{\frac{L}{g - \frac{qE}{m}}}}{2\pi \sqrt{\frac{L}{g}}} = \sqrt{\frac{g}{g - \frac{qE}{m}}} = \left(\frac{g}{g - \frac{qE}{m}}\right)^{1/2}$.

Case 3: Electric field is horizontal. The electric force is $F_E = qE$ horizontally. The gravitational force is $F_g = mg$ vertically downwards. The effective acceleration is $g_{eff} = \sqrt{g^2 + (\frac{qE}{m})^2}$. The time period is $T = 2\pi \sqrt{\frac{L}{g_{eff}}} = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + (\frac{qE}{m})^2}}}$. The ratio $T/T_0 = \frac{2\pi \sqrt{\frac{L}{\sqrt{g^2 + (\frac{qE}{m})^2}}}}{2\pi \sqrt{\frac{L}{g}}} = \sqrt{\frac{g}{\sqrt{g^2 + (\frac{qE}{m})^2}}} = \left(\frac{g}{\sqrt{g^2 + (\frac{qE}{m})^2}}\right)^{1/2}$.

Comparing the results with the given options, options (1) and (3) match the result obtained when the electric field is vertically downwards (assuming q is positive). Since options (1) and (3) are identical, it is likely that the intended problem involves a vertical downward electric field, despite the diagram showing vertical plates and thus a horizontal electric field. Or, the diagram is just a generic diagram for a pendulum between plates and the electric field direction is not implied by the diagram. Given the options, the scenario with a vertical electric field is the only one that yields a matching expression.

Let's assume the electric field is vertical and downwards. $T_0 = 2\pi \sqrt{\frac{L}{g}}$ $T = 2\pi \sqrt{\frac{L}{g + \frac{qE}{m}}}$ $\frac{T}{T_0} = \frac{2\pi \sqrt{\frac{L}{g + \frac{qE}{m}}}}{2\pi \sqrt{\frac{L}{g}}} = \sqrt{\frac{g}{g + \frac{qE}{m}}} = \left(\frac{g}{g + \frac{qE}{m}}\right)^{1/2}$

This matches options (1) and (3).

The final answer is $\boxed{(\frac{g}{g+\frac{qE}{m}})^{1/2}}$.

The final answer is $\boxed{(\frac{g}{g+\frac{qE}{m}})^{1/2}}$. The option (1) and (3) are identical. Let's choose (1) as the answer.

The final answer is $\boxed{(\frac{g}{g+\frac{qE}{m}})^{1/2}}$.