Question

Electric charges of $1\mu C,-1\mu C$ and $2\mu C$ are placed in air at the corners A, B and C respectively of an equilateral triangle ABC having length of each side 10 cm. The resultant force on the charge at C is $(\frac{1}{4\pi {ϵ}_0}=9\times {10}^{9}N-m^2/c^2)$:

• A. $0.9N$
• B. $1.8N$
• C. $2.7N$
• D. $3.6N$

Answer 1

The correct option is B $1.8N$

$\text{Step 1: Draw the diagram}$ $\text{[Ref. Fig.]}$

Mention the direction of all forces on charge at $C$ due to all other charges.

$\text{Step 2: Forces due to other charges at}$ $C$

Apply Coulomb's law.

$F_A=\frac{1}{4\pi {\in }_0}\frac{q_C{q}_A}{r_{AC}^2}=\frac{9\times {10}^9\times 2\times {10}^{-6}C\times {10}^{-6}C}{(10\times {10}^{-2}m{)}^2}=1.8N$

$F_B=\frac{1}{4\pi {\in }_0}\frac{q_C{q}_B}{r_{BC}^2}=\frac{9\times {10}^9\times 2\times {10}^{-6}C\times {10}^{-6}C}{(10\times {10}^{-2}m{)}^2}=1.8N$

$\text{Step 3: Resultant force at C}$

$F=\sqrt{F_A^2+F_B^2+2F_A{F}_B\cos {120}^{\circ }}$

$=\sqrt{(1.8{)}^2+(1.8{)}^2+2\times 1.8\times 1.8\times (-\frac{1}{2})}$

$=\sqrt{(1.8{)}^2+(1.8{)}^2-(1.8{)}^2}=1.8N$

Hence, Option(B) is correct.