Question

two point charges +6q and -8q are placed at the vertices 'B' and 'C' of an equilateral triangle abc of side 'a’. obtain the expression for the direction of the resultant electric field at the vertex A due to these two charges.

Answer 1

Given that,

Two point charges

Charge on B = +6q

Charge on C =-8q

We know that,

The electric field is

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{d^2}$

Where, q = charge

r = distance

We need to calculate the magnitude of electric field of $E_{BA}$

Using formula of electric field

$E_{BA}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q_{B}}{d^2}$

Put the value into the formula

$E_{BA}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{6q}{a^2}$

$E_{BA}=6E$

We need to calculate the magnitude of electric field of $E_{BC}$

Using formula of electric field

$E_{BC}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q_{C}}{d^2}$

Put the value into the formula

$E_{BC}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{8q}{a^2}$

$E_{BC}=8E$

We need to calculate the magnitude of the resultant field

Using formula for the resultant field

$E_{net}=\sqrt{E_{BA}^2+E_{AC}^2+2E_{BA}E_{AC}\cos120^{\circ}}$

Put the value into the formula

$E_{net}=\sqrt{(6E)^2+(8E)^2+2\times6E\times8E\times(-\dfrac{1}{2})}$

$E_{net}=E\sqrt{52}$

$E_{net}=\dfrac{1}{2}\times\dfrac{q\sqrt{52}}{a^2}$

(b). If the resultant field makes an angle β with AC

We need to calculate the direction of resultant electric field

Using formula of direction

$\tan\beta=\dfrac{E_{BA}\sin120^{\circ}}{E_{AC}+E_{BA}\cos120}$

$\tan\beta=\dfrac{6E\times\dfrac{\sqrt{3}}{2}}{8E+6E\dfrac{-1}{2}}$

$\tan\beta=\dfrac{3\sqrt{3}}{5}$

$\beta=\tan^{-1}(\dfrac{3\sqrt{3}}{5})$

Hence, (I). The magnitude of the resultant field is $\dfrac{1}{2}\times\dfrac{q\sqrt{52}}{a^2}$

(II).  The direction of resultant electric field is $\tan^{-1}(\dfrac{3\sqrt{3}}{5})$