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

Four-point charges $Q, q, Q$ and $q$ are placed at the corners of a square of side $' a'$ as shown in the figure.

Find the resultant electric force on a charge $Q$ .

Accepted Answer

According to the question,

From the diagram given above, we can see, there are three forces due to the three other charges are acting on the charge $Q$ kept at corner at one of the corners of the square.

Applying the coulomb's law, the force acting between two charges $q_{1}$ and $' a'$ , kept at a distance $r$ can be given as:

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

Using the coulomb's law to calculate forces acting between the charges kept at B and C, we get:

$F_{1} = \frac{1}{4 {πε}_{0}} \times \frac{q Q}{a^{2}}$ (along BC)

Using the coulomb's law to calculate forces acting between the charges kept at D and C, we get:

$F_{2} = \frac{1}{4 {πε}_{0}} \times \frac{q Q}{a^{2}}$ (along DC)

Using the coulomb's law to calculate forces acting between the charges kept at A and C, we get:

$F_{3} = \frac{1}{4 {πε}_{0}} \times \frac{Q^{2}}{{(a \sqrt{2})}^{2}}$

From the diagram, it can be seen that the forces $F_{1}$ and $F_{2}$ are perpendicular to each other. Thus, using the principle of superposition, the resultant force due to these two forces will be:

$F = \sqrt{{(F_{1})}^{2} + {(F_{2})}^{2} + 2 (F_{1}) (F_{2}) \cos (90 °)}$

Substitute, $\cos (90 °) = 0$

$F_{1} = F_{2}$ , we get:

$F = \sqrt{2 {(F_{1})}^{2}} = F_{1} \sqrt{2}$
$F = \frac{\sqrt{2}}{4 {πε}_{0}} \times \frac{q Q}{a^{2}}$ (Along AC)

Thus, the net force on the charge at D can be calculated using the superposition principle, we get:

$F' = F + F_{3}$

$F' = \frac{\sqrt{2}}{4 {πε}_{0}} \times \frac{q Q}{a^{2}} + \frac{1}{4 {πε}_{0}} \times \frac{q Q}{{(a \sqrt{2})}^{2}}$

$F' = \frac{Q}{4 {πε}_{0}} [\frac{\sqrt{2} q}{a^{2}} + \frac{Q}{2 a^{2}}]$

$F' = \frac{Q}{4 {πε}_{0}} [\frac{2 \sqrt{2} q + Q}{a^{2}}]$ (along AC)

Embibe Experts Solutions for Chapter: Electric Charges and Fields, Exercise 1: cbse-2020

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