Chapter 33, Problem 24A

A charged ball is suspended by a string in a uniform electric field pointing to the right. The string makes an angle $\theta$ with the vertical, as two forces act on the ball—one gravitational and the other electric.

Consider the following three scenarios for the ball’s mass and charge.

(A) $\text{mass}=3g;\text{charge}=2\text{nC}$

(B) $\text{mass}=6g;\text{charge}=8\text{nC}$

(C) $\text{mass}=9g;\text{charge}=4\text{nC}$

Rank, from greatest to least, the angle the string makes with the vertical.

To determine

To rank: The angle the string makes with the vertical from greatest to least.

Answer to Problem 24A

  $\left(B\right)>\left(A\right)>\left(C\right)$

Explanation of Solution

Introduction:

A force field that is surrounding an electric charge or a group of charges can be defined as an electric field. Mathematically, the electric field strength is given as:

  $E=\frac{F}{q}$

Hence, the electric field is strongest when the force is greatest and vice versa.

According to the statement, the two forces acting on the ball are: electric and gravitational. So, the gravitational force here is acting vertically downwards and given by: $F=mg$

While the electric force is acting horizontally rightwards and given by: $F=qE$

Hence, the angle with the vertical can be given by:

  $\tan \theta =\frac{qE}{mg}$

Using the above expression in the given three scenarios we get:

1. Mass is $m=3\text{ g}$ and the charge is $2\text{ nC}$:

$\tan \theta =\frac{2E}{3g}$

2. Mass is $m=6\text{ g}$ and the charge is $\text{8 nC}$:

$\tan \theta =\frac{8E}{6g}=\frac{4E}{3g}$

3. Mass is $m=9\text{ g}$ and the charge is $\text{4 nC}$:

$\tan \theta =\frac{4E}{9g}$

Hence, the ratio is higher in case (B), and lower in case (C). So, the ranking can be given as:

  $\left(B\right)>\left(A\right)>\left(C\right)$

Conclusion:

Hence, the ranking will be $\left(B\right)>\left(A\right)>\left(C\right)$