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Chapter 11: Problem 1550

A simple pendulum consists of a small sphere of mass \(\mathrm{m}\) suspended by a thread of length \(\ell\). The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength \(\mathrm{E}\) directed Vertically upwards. If the electrostatic force acting on the sphere is less than gravitational force the period of pendulum is (A) $\mathrm{T}=2 \pi[\ell /\\{\mathrm{g}-(\mathrm{q} \mathrm{E} / \mathrm{m})\\}]^{(1 / 2)}$ (B) \(\mathrm{T}=2 \pi(\ell / \mathrm{g})^{(1 / 2)}\) \(\left.\left.\left.\mathrm{m}_{\mathrm{}}\right\\}\right\\}\right]^{(1 / 2)}\) (D) \(\mathrm{T}=2 \pi[(\mathrm{m} \ell / \mathrm{qE})]^{(1 / 2)}\) (C) \(\mathrm{T}=2 \pi[\ell /\\{\mathrm{g}+(\mathrm{qE} / \mathrm{t}\)

Short Answer

Expert verified
The short answer to the question is: The period of the pendulum is given by (A): \(T = 2\pi\left[\frac{l}{\{g - (\frac{qE}{m})\}}\right]^{\frac{1}{2}}\).

Step by step solution

01

Analyze the forces acting on the pendulum

There are two forces acting on the pendulum sphere: the gravitational force (F_gravity) and the electrostatic force (F_electric). The gravitational force is given by: \[F_{gravity} = m*g\] where m is the mass of the sphere and g is the acceleration due to gravity. The electrostatic force can be determined using Coulomb's Law: \[F_{electric} = q*E\] where q is the charge on the sphere and E is the electric field strength. Given that the electrostatic force is less than the gravitational force, we have: \[F_{electric} < F_{gravity}\]
02

Determine the effective gravitational force acting on the pendulum

Since both forces are acting vertically, we can determine the effective gravitational force (F_eff) acting on the sphere. \[F_{eff} = F_{gravity} - F_{electric} = m*g - q*E\]
03

Calculate the period of the pendulum

For a simple pendulum, the period (T) is given by the equation: \[T = 2 \pi \sqrt{\frac{l}{g_{eff}}}\] Substituting the value of the effective gravitational force (F_eff) acting on the pendulum, we get: \[T = 2\pi\sqrt{\frac{l}{\frac{F_{eff}}{m}}}\] Substitute the value of \(F_{eff}\): \[T = 2\pi\sqrt{\frac{l}{\frac{m*g - q*E}{m}}}\] Simplify the equation: \[T = 2\pi\sqrt{\frac{l}{g - \frac{qE}{m}}}\] Comparing this result with the given options, the answer is (A): \[T = 2\pi[\frac{l}{\{g - (\frac{qE}{m})\}}]^{\frac{1}{2}}\]

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Most popular questions from this chapter

The electric flux for gaussian surface \(\mathrm{A}\) that enclose the $\ldots \ldots\( charged particles in free space is (given \)\left.q_{1}=-14 n c, q_{2}=78.85 \mathrm{nc}, q_{3}=-56 n c\right)$ (A) \(10^{4} \mathrm{Nm}^{2} / \mathrm{C}\) (B) \(10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (C) \(6.2 \times 10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (D) \(6.3 \times 10^{4} \mathrm{Nm}^{2} / \mathrm{C}\)

The electric potential \(\mathrm{V}\) at any point $\mathrm{x}, \mathrm{y}, \mathrm{z}\( (all in meter) in space is given by \)\mathrm{V}=4 \mathrm{x}^{2}$ volt. The electric field at the point \((1 \mathrm{~m}, 0,2 \mathrm{~m})\) in \(\mathrm{Vm}^{-1}\) is \((\mathrm{A})+8 \mathrm{i} \wedge\) (B) \(-8 \mathrm{i} \wedge\) (C) \(-16 \mathrm{i}\) (D) \(+16 \mathrm{i}\)

Iaking earth to be a metallic spheres, its capacity will approximately be (A) \(6.4 \times 10^{6} \mathrm{~F}\) (B) \(700 \mathrm{pF}\) (C) \(711 \mu \mathrm{F}\) (D) \(700 \mathrm{pF}\)

Two thin wire rings each having a radius \(R\) are placed at a distance \(d\) apart with their axes coinciding. The charges on the two rings are \(+q\) and \(-q\). The potential difference between the centers of the two rings is $\ldots .$ (A) 0 (B) $\left.\left[\mathrm{q} /\left(2 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{(}^{2}+\mathrm{d}^{2}\right)\right\\}\right]$ (C) $\left[\mathrm{q} /\left(4 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{\left. \left.\left(\mathrm{R}^{2}+\mathrm{d}^{2}\right)\right\\}\right]}\right.\right.$ (D) \(\left[(q R) /\left(4 \pi \epsilon_{0} d^{2}\right)\right]\)

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and work done by the battery will be (A) \((1 / 2)\) (B) \((2 / 1)\) (C) 1 (D) \((1 / 4)\)
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