Question

A simple pendulum consists of a small sphere of mass m suspended by a thread of length l. The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength e directed vertically downwards. Find the period of oscillation of the pendulum due to the electrostatic force acting on the sphere, neglecting the effect of the gravitational force.

Answer 1

Explanation:

A simple pendulum consists of a small sphere of mass m suspended by a thread of length l. The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength E directed vertically upwards.

Answer 2

Given:

• The sphere of mass m has charge q on it.
• The length of pendulum is l.
• The electric field E is directed vertically upwards.

To Find:

• The period of oscillation of the pendulum due to the electrostatic force acting on the sphere.

Solution:

Due to electric field, the sphere experiences a force in the upward direction.

   This force is given as $F = qE$

Due to this force, the acceleration produced in the upward direction in   sphere is

                       $a = \frac{F}{m}$

putting the value of $F$

                     $a = \frac{qE}{m}$

Force

             $F = ma = m \times \frac{qE}{m} = qE$

New force $F_(net) = mg + qE$

Net acceleration  $= \frac{F_(net)}{m}$

                   $a_(net) = \frac{mg + qE}{m}$

                  $a_(net) = g + \frac{ qE}{m}$

                             

The Time period of simple pendulum is thus given as

               $T = 2\pi \sqrt{\frac{l}{a} }$

Putting value of a

              $T = 2\pi \sqrt{\frac{l}{g + \frac{qE}{m} } }$

Thus,  The time period of oscillation of the pendulum $T = 2\pi \sqrt{\frac{l}{g + \frac{qE}{m} } }$

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