Question

three point charges of 2 microcoulombs -4 microcoulomb and 8 microcoulomb are placed at three vertices of an equilateral triangle of side length 10 centimetre the potential at the center of triangle is​

Answer 1

Three point charges of $2\mu C$, $-4\mu C$ and $8\mu C$ are placed at three vertices of an equilateral triangle of side length 10cm.

we have to find potential at the centre of triangle.

first find distance between charges and centre of triangle.i.e., OA , OB and OC

OA = OB = OC = $\frac{2}{3}\sqrt{(10)^2-5^2}=\frac{2}{3}5\sqrt{3}=\frac{10}{\sqrt{3}}cm$

now potential at the centre of triangle , V = $\frac{Kq_A}{OA}+\frac{kq_B}{OB}+\frac{kq_C}{OC}$

= $\frac{\sqrt{3}k(2\mu C)}{10cm}+\frac{\sqrt{3}k(-4\mu C)}{10cm}+\frac{\sqrt{3}k(8\mu C)}{10cm}$

= $\frac{6\mu C\sqrt{3}k}{10cm}$

= 6 × 10^-6 × 1.732 × 9 × 10^9/(0.1)

= 54 × 1.732 × 10³ N/C

= 93.528 × 10³ N/C

= 9.3528 × 10⁴ N/C