Question

Two point charges 3nC and -3nC are placed at A and B corners of an equilateral triangle of side 0.1m .find the resultant electric field at c

Answer 1

Given :

Two point charges 3nC and -3nC placed at the two corners of an equilateral triangle.

side of the equilateral triangle = 0.1 m

To Find :

Electric Field at C

Solution :

Electric Field produced by a charge of 3nC (E) = $\frac{1}{4\pi E_{0} }$ $\frac{q}{r^{2} }$

(Where, q = charge; r = distance between the point where we need to find the field and the charge)

                     E = $\frac{1}{4\pi E_{0} } \frac{3 X 10^{-9} }{(0.1)^{2} }$    (3nC = 3 × $10^{-9}$ C)

                     E =  $9 X 10^{9} X \frac{3 X 10^{-9} }{(0.1)^{2} }$

                     E = 2700 V/m

Resultant Electric Field at point C (E') = $\sqrt{E^{2} + E^{2} + 2.E.Ecos120}$

(angle between two field vectors is 120°)

                                                             = $\sqrt{2E^{2} + 2E^{2}sin30 }$ (cos120° = sin30°)

                                                             = $\sqrt{2E^{2} (1 + sin30)}$

                                                             = $\sqrt{2E^{2}(1 + \frac{1}{2} ) }$

                                                             = $\sqrt{2E^{2} X \frac{3}{2} }$

                                                             = $\sqrt{3} E$

                                                             = $\sqrt{3}$ × 2700

                                                            = 4676.54 V/m

∴ The resultant electric field is 4676.54 V/m.

Answer 2

$\underline\red{Answer:}$

$\red{ \small{\text{Electric Field produced by a charge: }}} \\ E = \frac{1}{4 π E_{0}} \frac{q}{ {r}^{2} }$

Where, q = charge; r = distance between the point.

$\blue{ \text{Electric field at c due to A is:}}$

$E_{A}=\frac{1}{4π E_{0}} \frac{q}{r²}$

$E_{A}=9×10^9×\frac{3×{10}^{ - 9} }{(0.1)²}$

$E_{A}=2700 N/c \: \: \: \text{ along Ac} \:$

$\pink{ \text{Electric field at c due to B \: is:}}$

$E_{B}=\frac{1}{4π E_{0}} \frac{q}{r²}$

$E_{B}=9×10^9×\frac{3×{10}^{ - 9} }{(0.1)²}$

$E_{A}=2700 N/c \: \: \: \text{ along AB} \:$

$\blue{\text{Resultant field at c}}$

$E=\sqrt{E_{A}^2+E_{B}²+2 E_{A}E_{B} cos \theta}$

$E_{A}=E_{B}=2700N/c$

Angle between them is 120°

$E=\sqrt{E_{A}^2+E_{A}²+{2}E_{A}E_{A} cos 120}$

$E=\sqrt{E_{A}^2+E_{A}²+\cancel{2}E_{A}²(- \frac{1}{ \cancel2}})$

$E=\sqrt{E_{A}^2+E_{A}² - E_{A}^{2} }$

$E=\sqrt{E_{A}²}$

$E=E_{A}$

$E=2700N/c$

∴ The resultant electric field is 2700N/c