A simple pendulum  of length L is placed between theplates of a parallal plate capacitor having electric field E, as shown in figure. its bob has mass m and charge q. The time period of the pendulum is given by:

  • Option 1)

    2\pi \sqrt{\frac{L}{\sqrt{g-\frac{qE}{m}}}}

     

  • Option 2)

    2\pi \sqrt{\frac{L}{\sqrt{g^{2}-\frac{q^{2}E^{2}}{m^{2}}}}}

     

  • Option 3)

    2\pi \sqrt{\frac{L}{g+\frac{qE}{m}}}

  • Option 4)

    2\pi \sqrt{\frac{L}{\sqrt{g^{2}+\left ( \frac{qE}{m} \right )^{2}}}}

Answers (1)

Two force act on the mass m

1) gravity force

2) force by electric field.

net force on m is =\sqrt{\left ( mg \right )^{2}+\left ( qE \right )^{2}}

So, net g is =\sqrt{ g ^{2}+\left ( qE \right/m )^{2}}

T=2\pi \sqrt{\frac{l}{g}}

=2\pi \sqrt{\frac{l}{\sqrt{g^{2}+\left ( \frac{qt}{m} \right )^{2}}}}


Option 1)

2\pi \sqrt{\frac{L}{\sqrt{g-\frac{qE}{m}}}}

 

Option 2)

2\pi \sqrt{\frac{L}{\sqrt{g^{2}-\frac{q^{2}E^{2}}{m^{2}}}}}

 

Option 3)

2\pi \sqrt{\frac{L}{g+\frac{qE}{m}}}

Option 4)

2\pi \sqrt{\frac{L}{\sqrt{g^{2}+\left ( \frac{qE}{m} \right )^{2}}}}

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