Electrostatics

Question: Two charges of $4\mu C$ each are placed at the corners A and B of an equilateral triangle of side length 0.2 m in air. The electric potential at C is $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\frac{{N{\rm{ - }}{m^2}}}{{{C^2}}}} \right]$



1) $9 \times {10^4}$V
2) $18 \times {10^4}$V
3) $36 \times {10^4}$V
4) $36 \times {10^{ - 4}}$V
Solution: Explanation: Explanation
electric-field-potential

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