Question

Two positive charge 4×10^-6C and 6×10^-6c are placed at the corners Aand B of an equilateral triangle of side 3m .Find the electric intensity of the corner C

Answer 1

✪ᴀɴsᴡᴇʀ✪

• $\red{\boxed{E_C= 2000\sqrt{13 + 6\sqrt{3}}\:N/C}}$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

☆ɢɪᴠᴇɴ☆

• Two positive charge are placed at corners A and B of equilateral triangle ABC.

• $Q_A = 4\times 10^{-6}C,$

• $Q_B = 6\times 10^{-6}C.$

• Side length of equilateral triangle, a = 3 m.

☆ᴛᴏ ғɪɴᴅ☆

• Electric Field Intensity at C.

☆ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ☆

Electric Field Intensity at C due to charge particle at A is given by

$\to \vec{E}_{CA} = \frac{KQ_A}{a^2}\hat{AC}$

Electric Field Intensity at C due to charge particle at B is given by

$\to \vec{E}_{BA} = \frac{KQ_B}{a^2}\hat{BC}$

Since ABC is equilateral, angle between $\hat{AC}$ and $\hat{BC}$ i. e. $\theta = 60^o$. Therefore,

$\:\:\:\:\:\:\: \vec{E}_C = \vec{E}_{CA} + \vec{E}_{BA}$

$\implies |\vec{E}_C|= |\vec{E}_{CA} + \vec{E}_{BA}|$

$\implies E_C=$

$\sqrt{({E_{CA})}^2 + {(E_{BA})}^2 + 2E_{CA}.E_{BA}cos(\theta)}$

$\sqrt{{(\frac{KQ_A}{a^2})}^2 + {(\frac{KQ_B}{a^2})}^2 + 2(\frac{KQ_A}{a^2}).(\frac{KQ_B}{a^2})cos(30^o)}$

$\frac{K}{a^2}\sqrt{{(Q_A)}^2 + {(Q_B)}^2 + 2(Q_A).(Q_B)cos(30^o)}$

$\frac{K}{a^2}\sqrt{{(4\times 10^{-6})}^2 + {(6\times 10^{-6})}^2 + 2(4\times 10^{-6}).(6\times 10^{-6})\frac{\sqrt{3}}{2}}$

$\frac{10^{-6}K}{a^2}\sqrt{4^2 + 6^2 + 4\times 6\sqrt{3}}$

$\frac{10^{-6}(9\times 10^9)}{3^2}\sqrt{52+ 24\sqrt{3}}$

$\frac{9\times 10^3}{9}\sqrt{4(13+ 6\sqrt{3})}$

$\implies E_C= 2\times 10^3\sqrt{13 + 6\sqrt{3}}$

$\implies E_C= 2000\sqrt{13 + 6\sqrt{3}}\:N/C$