Question

In an electricity experiment, a 1.0g plastic ball is suspended on a60-cm-long string and given an electric charge. A charged rodbrought near the ball exerts a horizontal electrical forceFelec on it, causing the ball to swing out to a 20deg.angle and remain there.

a. What is the magnitude of Felec ?
b.What is the tension in the string?

Answer 1

a) Magnitude of the electric force: $3.57\cdot 10^{-3} N$

b) Tension in the string: 0.010 N

Explanation:

a)

When the charged rod is brought near the ball, then the ball remains "suspended" in an inclined position. Therefore, we can analzye the forces acting in two perpendicular directions:

- Along the horizontal direction, we have the electric force $F_E$, pushing in one direction, and the component of the tension in the string acting in the opposite direction, $T sin \theta$, where T is the tension and $\theta=20^{\circ}$ is the angle with the vertical

- Along the vertical direction, we have the weight of the ball, $mg$, acting downward (where $m=1.0 g = 0.001 kg$ is the mass of the ball and $g=9.8 m/s^2$ is the acceleration due to gravity), and the component of the tension acting in the upward direction, $T cos \theta$

Therefore, since the ball is in equilibrium, we have the two equations:

$T sin \theta =F_E\\Tcos \theta = mg$

By dividing the two equations, we get

$tan \theta=\frac{F_E}{mg}$

an solving for the electric force, we find

$F_E=mg tan \theta=(0.001)(9.8)tan 20^{\circ}=3.57\cdot 10^{-3} N$

b)

The tension in the string can now be found by using either of the two equations above; for instance, by using the equation along the horizontal direction,

$T sin \theta =F_E$

Where

$F_E=3.57\cdot 10^{-3} N$ is the electric force

$\theta=20^{\circ}$ is the angle with the vertical

We find the tension in the string:

$T=\frac{F_E}{sin \theta}=\frac{3.57\cdot 10^{-3} N}{sin 20^{\circ}}=0.010 N$

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Answer 2

The tension in the string is due to the electric force, and the weight of the ball.

The correct responses are;

a. The magnitude of $F_{elec}$ is approximately 3.57 × 10⁻³ N

b. The tension in the string is approximately 1.044 × 10⁻² N

Reasons:

The given  parameters are;

Mass of the plastic ball = 1.0 g

Length of the string from which the ball is suspended = 60 cm

Angle of swing of the ball, θ = 20°

a. The magnitude of the electric force, $F_{elec}$ is given as follows;

$F_{elec}$ = T×sin(20°)

W = T×cos(20°)

Where;

T = Tension in the string

W = The weight of the ball = 1.0 g × 9.81 m/s² = 0.00981 N

$\dfrac{F_{elec}}{W} = \dfrac{T \times sin(20^{\circ}}{T \times cos(20^{\circ}} = tan(20^{\circ})$

$F_{elec}$ = W × tan(20°)

∴ $F_{elec}$ = 0.0098 N × tan(20°) ≈ 0.00357 N

The magnitude of $F_{elec}$ ≈ 0.00357 N = 3.57 × 10⁻³ N

b. The tension in the string is found from W = T×cos(θ)

$T = \dfrac{0.00981 \ N}{cos( 20^{\circ})} \approx 0.01044 \, N$

The tension in the string ≈ 0.01044 N = 1.044 × 10⁻² N

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