Question

Four charges are placed at the circumference of a dial clock as shown in the figure. If the clock has only hour hand, then the resultant force on a charge $+q_0$ placed at the centre, points in the direction which shows the time as

• A. $1:30$
• B. $7:30$
• C. $4:30$
• D. $10:30$

Answer 1

The correct option is B $7:30$

Let $r$ be the radius of the dial clock.

Force due to each charge on $q_0$ can be considered accordding to Coulombs law as,
$$F=\frac{k{q}_1{q}_2}{r^2}$$
As we know, like charges repel each other and unlike charges attract each other.

The positive charges on the circumference of dial clock repel the charge $+q_0$ present at the centre whereas the negative charges attract the charge $+q_0$ as shown in the force diagram given below.


From the diagram we can deduce that,

Net force on $+q_0$ by diametrically opposite charges is $2F$.


Hence, Resultant force $F^{\prime }=\sqrt{{\left(2F\right)}^2+{\left(2F\right)}^2}$
($\because \theta ={90}^{\circ }$)
$\Rightarrow F^{\prime }=2\sqrt{2}F$

Direction of the resultant force is given by ,
$\tan \theta =\left(\frac{2F}{2F}\right)=1\Rightarrow \theta ={45}^{\circ }$ (with horizontal)

Since, the time difference in each quarter is $3\text{hrs}$.

The resultant force will be along the direction shown by time $7:30$.

Hence, option (b) is the correct answer.