Four identical point charges (+ 2.50 nC)are placed at the corners of a rectangle which measures 2.00 m * 4.00 m.If the electric potential is taken to be zero at infinity,what is the potential at the geometric center of this rectangle?

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Quizplus · Accepted Answer

The electric potential $V$ at a point due to a point charge $q$ is given by $V = \frac{kq}{r}$, where $k$ is Coulomb's constant ($9 \times 10^9 \, \text{Nm}^2/\text{C}^2$) and $r$ is the distance from the charge to the point in question. For multiple charges, the total potential is the algebraic sum of the potentials due to each charge because potential is a scalar quantity.Given that each charge is $+2.50 \, \text{nC} = 2.50 \times 10^{-9} \, \text{C}$, and they are placed at the corners of a rectangle with dimensions $2.00 \, \text{m} \times 4.00 \, \text{m}$, the distance from each charge to the geometric center can be found using the Pythagorean theorem. The geometric center is at the midpoint of the rectangle, so the distance from each corner to the center is $\sqrt{(2.00/2)^2 + (4.00/2)^2} = \sqrt{1^2 + 2^2} = \sqrt{5} \, \text{m}$.Thus, the potential at the center due to one charge is $V = \frac{(9 \times 10^9)(2.50 \times 10^{-9})}{\sqrt{5}}$. Since there are four identical charges, the total potential is four times this value:$$V_{\text{total}} = 4 \times \frac{(9 \times 10^9)(2.50 \times 10^{-9})}{\sqrt{5}} = \frac{90 \times 10^0}{\sqrt{5}}$$$$V_{\text{total}} = \frac{90}{\sqrt{5}} = 90 \times \frac{\sqrt{5}}{5} = 18 \times \sqrt{5}$$Given that $\sqrt{5}$ is approximately $2.236$, the total potential is approximately $18 \times 2.236 = 40.248 \, \text{V}$, which rounds to $40.2 \, \text{V}$, making option E the correct answer.

Four Identical Point Charges (+ 2