A $2.00 \mathrm{nC}$ charge is placed on the $\mathrm{x}$-axis at $\mathrm{x}=-3.00 \mathrm{~cm}$, another $2.00 \mathrm{nC}$ charge is placed on the $\mathrm{x}$-axis at $\mathrm{x}=3.00 \mathrm{~cm}$, and a third $2.00 \mathrm{nC}$ charge is placed on the $\mathrm{y}$-axis at $\mathrm{y}=3.00 \mathrm{~cm}$. What is the potential at the origin?

Question

Quizplus · Accepted Answer

The potential at the origin due to each charge is calculated using $ V = \frac{kq}{r} $. Each charge contributes equally, and the total potential is the sum of potentials from each charge. The distance $ r $ for each charge from the origin is $ 3 \mathrm{~cm} $ or $ 0.03 \mathrm{~m} $. Thus, $ V = 3 \times \frac{(9 \times 10^9 \mathrm{~Nm^2/C^2})(2 \times 10^{-9} \mathrm{~C})}{0.03 \mathrm{~m}} = 1800 \mathrm{~V} $.

A 2.00nC2.00 \mathrm{nC}2.00nC Charge Is Placed on The x\mathrm{x}x -Axis At x=−3.00 cm\mathrm{x}=-3.00 \mathrm{~cm}x=−3.00 cm

A $2.00 \mathrm{nC}$ Charge Is Placed on The $\mathrm{x}$ -Axis At $\mathrm{x}=-3.00 \mathrm{~cm}$