Question

A particle with a charge of 2.0 × 10−4 C is placed directly below and at a separation of 10 cm from the bob of a simple pendulum at rest. The mass of the bob is 100 g. What charge should the bob be given so that the string becomes loose?

Answer 1

Charge = $\begin{equation}0.0544 \times 10^{-9} \mathrm{N}\end$  should the bob be given so that the string becomes loose.

Explanation:

Given data in the question :        

Bob’s mass = 100 g = 0.1 kg  

So, in the string tension = 0.1×9.8 =0.98 N

A charged particle $\begin{equation}\left(q_{1}\right)=2.0 \times 10^{-4} \mathrm{C}\end$

The first bob should be repelled if the tension is 0.

$\begin{equation}F=\frac{k q_{1} q_{2}}{r^{2}}\end$

Charge = $\begin{equation}0.0544 \times 10^{-9} \mathrm{N}\end$  should the bob be given so that the string becomes loose.

Explanation:

Given data in the question :        

Bob’s mass = 100 g = 0.1 kg  

So, in the string tension = 0.1×9.8 =0.98 N

A charged particle $\begin{equation}\left(q_{1}\right)=2.0 \times 10^{-4} \mathrm{C}\end$

The first bob should be repelled if the tension is 0.

$\begin{equation}F=\frac{k q_{1} q_{2}}{r^{2}}\end$

T-mg+F = 0

T = mg-F

T = mg  

$\begin{equation}0.98=\frac{9 \times 10^{9} \times 2 \times 10^{-4} \times q_{2}}{0.01^{2}}\end$

$\begin{equation}0.98=\frac{18 \times 10^{5} \times q_{2}}{0.01^{2}}\end$

$\begin{equation}0.98=\frac{18 \times 10^{5} \times q_{2}}{1 \times 10^{-4}}\end$

$\begin{equation}0.98 \times 10^{-4}=18 \times 10^{5} \times q_{2}\end$

$\begin{equation}q_{2}=\frac{0.98 \times 10^{-4}}{18 \times 10^{5}}\end$

$\begin{equation}q_{2}=\frac{0.98 \times 10^{-9}}{18}\end$

$\begin{equation}q_{2}=0.0544 \times 10^{-9} \mathrm{N}\end$

Therefore, the charge is $\begin{equation}0.0544 \times 10^{-9} \mathrm{N}\end$.

$\begin{equation}0.98=\frac{9 \times 10^{9} \times 2 \times 10^{-4} \times q_{2}}{0.01^{2}}\end$

$\begin{equation}0.98=\frac{18 \times 10^{5} \times q_{2}}{0.01^{2}}\end$

$\begin{equation}0.98=\frac{18 \times 10^{5} \times q_{2}}{1 \times 10^{-4}}\end$

$\begin{equation}0.98 \times 10^{-4}=18 \times 10^{5} \times q_{2}\end$

$\begin{equation}q_{2}=\frac{0.98 \times 10^{-4}}{18 \times 10^{5}}\end$

$\begin{equation}q_{2}=\frac{0.98 \times 10^{-9}}{18}\end$

$\begin{equation}q_{2}=0.0544 \times 10^{-9} \mathrm{N}\end$

Therefore, the charge is $\begin{equation}0.0544 \times 10^{-9} \mathrm{N}\end$.

Answer 2

Change on the bob should be 0.054 * 10^-9 N

Explanation:

Mass of the bob = 100 g = 0.1 kg

So Tension in the string = 0.1 × 9.8 = 0.98 N

For the Tension to be 0, the charge below should repel the first bob.

F = kq1q2 / r²

T - mg + F = 0  implies that T = mg - f, so T = mg

0.98 = 9 * 10^9 * 2 * 10^-4 * q2 / 0.01²

q2 = (0.98 * 1 * 10^-2 )  / (9 * 2 * 10^5)

 = 0.054 * 10^-9 N