Name: Electric and Gravitational Forces
Uploaded: 2019-07-19
Duration: 3 min 11 s
Description: Do you know the difference and similarities between the gravitational force and electrostatic force? Watch this video to explore it. Don't miss it.

Question

Two point charges $+ q$ and $- 2 q$ are placed at the vertices ' $B$ ' and ' $C$ ’ of an equilateral triangle $A B C$ of side ‘ $a$ ’ as given in the figure. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at the vertex $A$ due to these two charges.

Accepted Answer

Given,

Electric field due to charge $q$ at point $A$ is ${\vec{E}}_{1} = \frac{1}{4 {πε}_{\circ}} \frac{q}{a^{2}}$ along $B A$

Electric field due to $- 2 q$ at point $A$ is ${\vec{E}}_{2} = \frac{1}{4 {πε}_{\circ}} \frac{2 q}{a^{2}}$ along $A C$

Angle between ${\vec{E}}_{1}$ and ${\vec{E}}_{2}$ $θ = 120^{\circ}$

(i) Resultant electric field at point $A$ is $E = \sqrt{E_{1}^{2} + E_{2}^{2} + 2 E_{1} E_{2} cosθ}$

$= \frac{1}{4 {πε}_{\circ}} \frac{q}{a^{2}} \sqrt{1^{2} + 2^{2} + 2 \times 1 \times 2 \cos 120^{\circ}}$

$= \frac{1}{4 {πε}_{\circ}} \frac{q}{a^{2}} \sqrt{3}$

(ii) Direction of resultant will make angle $α$ with $AC$ so that,

$\tan α = \frac{E_{1} \sin θ}{E_{2} + E_{1} \cos θ} = \frac{\frac{1}{4 {πε}_{0}} \frac{q}{a^{2}} \sin 120^{\circ}}{\frac{1}{4 {πε}_{0}} \frac{2 q}{a^{2}} + \frac{1}{4 {πε}_{0}} \frac{q}{a^{2}} \cos 120^{\circ}} = \frac{\frac{\sqrt{3}}{2}}{2 - \frac{1}{2}} = \frac{1}{\sqrt{3}}$

$\Rightarrow α = 30^{\circ}$

The direction of the resultant will make angle $30 °$ with $A C$ .

Important Questions on Electric Charges and Field

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Two point charges +q and -2q are placed at the vertices 'B' and 'C’ of an equilateral triangle ABC of side ‘a’ as given in the figure. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at the vertex A due to these two charges.

Important Questions on Electric Charges and Field

Learn and Prepare for any exam you want !