The bob of a simple pendulum has mass 2g and a charge of 5.0 μ C. It is at rest in a uniform horizontal electric field of intensity 2000 V/m. At equilibrium, the angle that the pendulum makes with the vertical is: (take g = 10 m/s2)

Question

The bob of a simple pendulum has mass 2g and a charge of 5.0 μ C. It is at rest in a uniform horizontal electric field of intensity 2000 V/m. At equilibrium, the angle that the pendulum makes with the vertical is: (take g = 10 m/s2) (A)  tan-1(5.0) (B)  tan-1(2.0) (C)  tan-1(0.5) (D)  tan-1(0.2)

Tardigrade Admin · Accepted Answer

tan θ = (qE/mg) = (5×10-6 ×2000/2×10-3 ×10) tanθ = (1/2) ⇒ θ = tan-1 (0.5)

Q. The bob of a simple pendulum has mass $2 g$ and a charge of $5.0 μ C$ . It is at rest in a uniform horizontal electric field of intensity $2000 V / m$ .
At equilibrium, the angle that the pendulum makes with the vertical is : (take $g = 10 m / s^{2}$ )

$tan^{- 1} (5.0)$

$tan^{- 1} (2.0)$

$tan^{- 1} (0.5)$

$tan^{- 1} (0.2)$

$tan θ = \frac{qE}{m g} = \frac{5 \times 1 0 ^{- 6} \times 2000}{2 \times 1 0 ^{- 3} \times 10}$
$tan θ = \frac{1}{2} \Rightarrow θ = tan^{- 1} (0.5)$

Q. The bob of a simple pendulum has mass 2g and a charge of 5.0μC. It is at rest in a uniform horizontal electric field of intensity 2000V/m. At equilibrium, the angle that the pendulum makes with the vertical is : (take g=10m/s2)

tan−1(5.0)

tan−1(2.0)

tan−1(0.5)

tan−1(0.2)