Question

The Δ ABC is defined by AB = AC = 40 cm. And angle at A is 30°. Two charges, each of magnitude 2 × 10 –6 C but opposite in sign, are placed at B and C, as shown in Fig. 15.9. Calculate the magnitude and direction of the field at A.

Answer 1

Answer:

Explanation:

Refer Attachment or figure.

Here, since both have same magnitude and the length of side of triangle (r) is also same. So the magnitude of electric field individually would also be the same. Let it be denoted by E.

Positive charge have outwards E and negative charges have E moving towards them. (refer figure)

Angle A = 30°

So, exterior angle A = 150° = ∅

Resultant = $\sf\sqrt{E^2+E^2+2(E)(E)cos\theta}$

Resultant = $\sf\sqrt{2E^2+2E^2cos\theta}$

Resultant = $\sf\sqrt{2E^2 + 2E^2(-\frac{\sqrt{3}}{2})}$

Resultant = $\sf\sqrt{2E^2 -\sqrt{3}E^2}$

Resultant = $\sf E\sqrt{2-\sqrt{3}}$

Now, E = $\sf\frac{kQ}{r^2}$

So, E = $\sf\frac{9\times 10^9\times2\times10^{-6}}{(4\times10^{-1})^2}\\\\\implies \frac{18\times10^3}{16\times10^{-2}}\\\\\implies 1.125 \times 10^5$

So resultant = 1.125 × 10⁵× √(2 - √3) N/C

Now, the angle it subtend with AC would be the half of 150°. (If two vectors are equal and subtend an angle ∅, then the resultant of the vector subtend angle ∅/2 with either of the vectors)

So, the direction of the field would be 75° with AC or 105° with AB in the direction of the negative charge.