CBSE Class 12 Physics 2024: question 17(b)

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. The centroid O is equidistant from all three vertices: $r=\frac{a}{\sqrt{3}}=\frac{0.1}{\sqrt{3}}$ m, so $r^2=\frac{0.01}{3}{\text{m}}^2$.

   Each field has magnitude
   $E=\frac{kq}{r^2}=\frac{9\times 10^9\times 10^{-12}\times 3}{0.01}=2.7{\text{N C}}^{-1}$.

   So $E_{OA}=E_{OB}=E_{OC}=2.7{\text{N C}}^{-1}$.

   1 mark

2. The three fields point away from their vertices and are at $120^{\circ }$ to each other.

   Resultant of $E_{OB}$ and $E_{OC}$:
   $E_{BC}=\sqrt{E^2+E^2+2E^2\cos 120^{\circ }}=E=2.7{\text{N C}}^{-1}$.

   This resultant is directed opposite to ${\vec{E}}_{OA}$. Therefore ${\vec{E}}_O={\vec{E}}_{OA}+{\vec{E}}_{BC}=0$.

   (Equivalently, three equal vectors at $120^{\circ }$ form a closed triangle, so their sum is zero.)

   1 mark

Answer: Net electric field at the centroid = 0

Revise the chapter

• Chapter: Electric Charges and Fields (notes, formulas, where marks are lost and every past question)