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: Net downward force
Weight $mg$ and electric force $qE$ both act downward, so the net force is $mg+qE$.

Step 2: Effective acceleration
Dividing by $m$ gives $g+\dfrac{qE}{m}$, which is larger than $g$, so the period must be shorter than $2\pi\sqrt{l/g}$.

Step 3: Compare the options
Option (D) ignores the field, option (A) has the field opposing gravity and option (C) needs a horizontal field. Option (B) matches.

Final Answer:
The period uses g plus Eq over m as the effective gravity. \[ \boxed{\text{(B) }2\pi \sqrt{\dfrac{l}{g+\dfrac{Eq}{m}}}} \]