Question

A simple pendulum having length \(l\) and mass 'm' is suspended between two plates having uniform electric field 'E' as shown in figure. The bob is given a charge \(q\). The time period 'T' of its vibration is

• A. \(2\pi \sqrt{\frac{l}{g-\frac{Eq}{m}}}\)
• B. \(2\pi \sqrt{\frac{l}{g+\frac{Eq}{m}}}\)
• C. \(2\pi \sqrt{\frac{l}{\left[g^{2}-\left(\frac{Eq}{m}\right)^{2}\right]^{1/2}}}\)
• D. \(2\pi \sqrt{\frac{l}{g}}\)

Hint:

The electric force on the charged bob adds to its weight, so the effective gravity is g plus Eq over m. Put this into the period formula.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A charged bob in a uniform field feels an extra force \(qE\) besides its weight. The pendulum then oscillates about the direction of the net force with an effective gravity.

Step 2: Direction of the force
In the arrangement shown, the electric force on the bob points downward, the same direction as gravity, so the two forces add.

Step 3: Effective gravity
\(g_{eff}=g+\dfrac{qE}{m}\)

Step 4: Time period
\(T=2\pi\sqrt{\dfrac{l}{g+\frac{Eq}{m}}}\)

Final Answer:
The time period is 2 pi root of l over g plus Eq over m. \[ \boxed{\text{(B) }2\pi \sqrt{\dfrac{l}{g+\dfrac{Eq}{m}}}} \]