Question

In an electricity experiment, a $1.0g$plastic ball is suspended on a $60$-cm-long string and given an electric charge. A charged rod brought near the ball exerts a horizontal electrical force $\overrightarrow{F_{elec}}$on it, causing the ball to swing out to a $20°$angle and remain there.

a. What is the magnitude of $\overrightarrow{F_{elec}}$?

b. What is the tension in the string

Short answer

(a) Magnitude of electrical force $\overrightarrow{F_{elec}}$= $0.0035N$.

(b) Tension in the string =$0.010N$.

Step-by-step solution

(a) Magnitude : 

The greatest extent of size and direction of an item is defined as magnitude.

(a) Explanation :

Plastic ball mass =$1.0g$, string's length =$60cm$and the string's angle with the vertical = $20°$.

The arrangement's free body diagram is illustrated below.

The force equation in the vertical direction at equilibrium is,

$$\begin{gathered} T\cos \theta -mg=0 \\ T=\frac{mg}{\cos \theta } \end{gathered}$$

$T$= tension of string,

$m$= plastic ball mass,

$\theta$= string's angle with the vertical,

$g$= acceleration due to gravity.

The force equation in the horizontal direction at equilibrium is,

$$\begin{gathered} F_{elec}-T\sin \theta =0 \\ {F}_{elec}=T\sin \theta \end{gathered}$$

$F_{elec}$= magnitude of electric force.

Substitute $\frac{mg}{\cos \theta }$for $T$.

$$\begin{gathered} F_{elec}=\frac{mg}{\cos \theta }\sin \theta \\ =mg\tan \theta \end{gathered}$$

Substitute $1.0g$for $m$, $9.81m/s^2$for $g$and $20°$for $\theta$to find $\overrightarrow{F_{elec}}$.

$$\begin{gathered} F_{elec}=\left(1.0g\times \frac{{10}^{-3}kg}{1g}\right)\left(9.81m/s^2\right)\tan 20° \\ =0.0035kg·m/s^2\times \frac{1N}{1kg·m/s^2} \\ =0.0035N \end{gathered}$$

Hence,$F_{elec}$magnitude is$0.0035N$.

(b) Tension : 

The pulling force transferred axially by a string, cable, chain, or similar item, or by each end of a rod, truss member, or similar three-dimensional object is known as tension.

(b) Explanation : 

Plastic ball mass =$1.0g$, string's length =$60cm$and the string's angle with the vertical = $20°$.

Substitute $1.0g$for $m$, $9.81m/s^2$for $g$and $20°$for $\theta$in equation to find T.

$$\begin{gathered} T=\frac{mg}{\cos \theta } \\ T=\frac{\left(1.0g\times {\displaystyle \frac{{10}^{-3}kg}{1g}}\right)\left(9.81m/s^2\right)}{\cos 20°} \\ =\frac{\left(0.00981kg·m/s^2\right)}{0.939} \\ =0.010kg·m/s^2\times \frac{1N}{1kg·m/s^2} \\ =0.010N \end{gathered}$$

Hence, string's tension is$0.010N$.