Question

Three charges \(+20\,\mu C\), \(+20\,\mu C\) and \(-20\,\mu C\) are placed at the vertices \(A\), \(B\) and \(C\) respectively of an equilateral triangle of side \(1\,m\). Find the net force acting on the charge at vertex \(A\).

• A. \(3600\sqrt3\,N\)
• B. \(3600\,N\)
• C. \(7200\,N\)
• D. \(1800\,N\)

Hint:

Always solve Coulomb force problems in two steps: \[ \boxed{\text{Step 1: Calculate each force using Coulomb's law}} \] \[ \boxed{\text{Step 2: Add the forces vectorially}} \] In an equilateral triangle, remember that the angle between two sides is \[ \boxed{60^\circ.} \]

Answer 1

The Correct Option is A

Concept: According to Coulomb's law, \[ \boxed{F=\frac{1}{4\pi\varepsilon_0}\frac{|q_1q_2|}{r^2}} \] where \[ k=\frac1{4\pi\varepsilon_0}=9\times10^9\,Nm^2/C^2. \] The resultant force is obtained by vector addition. For an equilateral triangle, \[ \angle BAC=60^\circ. \]

Step 1: Write the given data.
Given, \[ q_A=+20\,\mu C, \] \[ q_B=+20\,\mu C, \] \[ q_C=-20\,\mu C, \] \[ r=1\,m. \]

Step 2: Calculate the force due to each charge.
Using Coulomb's law, \[ F = 9\times10^9 \times \frac{(20\times10^{-6})^2}{1^2}. \] Since \[ (20\times10^{-6})^2 = 400\times10^{-12}, \] therefore, \[ F = 9\times10^9 \times 400\times10^{-12} = 3.6\,N. \] Thus, \[ \boxed{F_{AB}=F_{AC}=3.6\,N.} \] The force due to \(B\) is repulsive, whereas the force due to \(C\) is attractive. Hence, the angle between the two forces is \[ 120^\circ. \]

Step 3: Find the resultant force.
Using the cosine rule, \[ R = \sqrt{F^2+F^2+2F^2\cos120^\circ}. \] Since \[ \cos120^\circ=-\frac12, \] we get \[ R = \sqrt{2F^2-F^2} = F. \] Thus, \[ \boxed{R=3.6\,N.} \]

Important Observation: Using the given values, \[ \boxed{R=3.6\,N.} \] Hence none of the given options match. The options (\(3600\sqrt3\), \(3600\), etc.) would only be obtained if the charges were in milli-coulombs (mC) or if the numerical data were different. Therefore, the question contains an error.