Name: Forces Between Multiple Charges
Uploaded: 2019-07-19
Duration: 6 min 57 s
Description: To find the force on a charge due to multiple charges by applying the superposition principle, we need to know the basic concepts first. Watch the video to l...

Question

A clock face has negative charges

- q, - 2 q, - 3 q, . . . . . - 12 q

fixed at the position of the corresponding numericals on the dial. The clock hands do not disturb the net field due to the point charges. At what time does the hour hand point in the direction of the electric field at the centre of the dial?

Accepted Answer

We know that,

$E = \frac{k q}{r^{2}}$ where $k = \frac{1}{4 π ε_{0}}$ , $q$ is the magnitude of charge and $r$ is the distance from the charge.

Since, $k$ is constant and $r$ is same for all the charges, we can say that $E \propto q$ .

We also know that, the direction of electric field is from positive to negative. Therefore, between two negative charges, the field will direct towards the charge having a higher magnitude. So, between $- 6 q$ and $- 12 q$ , the direction of field $E_{1}$ will be towards $- 12 q$ as shown in the figure. Similarly, the direction of fields $E_{2}, E_{3}, E_{4}, E_{5}$ and $E_{6}$ are also shown in the figure.

Since, the difference between two diametrically opposite charges is same $(- 6 q)$ and the distance is also same, we can say that for all such charges, the magnitude of field is same. Let's say $E$ .

$∴ E_{1} = E_{2} = E_{3} = E_{4} = E_{5} = E_{6} = E$ .

Now, the angle between each charge would be,

$\frac{360}{12} = 30 °$

Now, we know that if the magnitude of two vectors and their resultant is same then the angle between them is $120 °$ .

Starting from $E_{1}$ , the other vector for which the difference in angle is $120 °$ with $E_{1}$ would be $E_{5}$ . The resultant of these vectors would be in the direction of $E_{3}$ having a magnitude of $2 E$ .

Similarly, for $E_{2}$ , the other vector for which the angle between them is $120 °$ would be $E_{6}$ . The resultant of these vectors would be in the direction of $E_{4}$ whose magnitude is $2 E$ .

Therefore, the net electric field at the center of the clock would be in the direction of the resultant of fields $E_{3}$ and $E_{4}$ .

Since, the magnitude of $E_{3}$ and $E_{4}$ is same, the direction of field would lie at the bisector between these two fields.

Since $E_{3}$ and $E_{4}$ represents $- 10 q$ and $- 9 q$ on the clock, their bisector would indicate a time of $9 : 30$ .

Important Questions on Electric Charges and Fields

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A clock face has negative charges -q, -2q, -3q, .....-12q fixed at the position of the corresponding numericals on the dial. The clock hands do not disturb the net field due to the point charges. At what time does the hour hand point in the direction of the electric field at the centre of the dial?

Important Questions on Electric Charges and Fields

Learn and Prepare for any exam you want !