Question

A charge of \(+2.0 \mathrm{mC}\) is located at \(x=0, y=0\) and a charge of $-4.0 \mathrm{mC}\( is located at \)x=0, y=3.0 \mathrm{m} .$ What is the electric potential due to these charges at a point with coordinates $x=4.0 \mathrm{m}, y=0 ?$

Short answer

Answer: The electric potential at point (4, 0) due to these charges is -2.697 x 10^6 V.

Step-by-step solution

Find the distance between each charge and the required point

To find the distance between each charge and the required point, we can use the distance formula: \(r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). For charge \(Q_1\), the distance between the point \((4,0)\) and \((0,0)\) is: \(r_1=\sqrt{(4-0)^2+(0-0)^2} = 4\,\mathrm{m}\) For charge \(Q_2\), the distance between the point \((4,0)\) and \((0,3)\) is: \(r_2=\sqrt{(4-0)^2+(0-3)^2} = 5\,\mathrm{m}\)

Calculate the electric potential due to each charge

Using the electric potential formula \(V_k = \frac{kQ}{r}\) and the electrostatic constant \(k = 8.99\times10^9 \mathrm{N m^2 C^{-2}}\), we can find the electric potential due to each charge at the point \((4,0)\). For charge \(Q_1\): \(V_1 = \frac{(8.99\times10^9)(2.0\times10^{-3})}{4} = 4.495\times10^6\,\mathrm{V}\) For charge \(Q_2\): \(V_2 = \frac{(8.99\times10^9)(-4.0\times10^{-3})}{5} = -7.192\times10^6\,\mathrm{V}\)

Calculate the net electric potential at the required point

Using the principle of superposition, the net electric potential at the required point is the sum of the electric potentials due to both charges. \(V_\mathrm{net} = V_1 + V_2 = 4.495\times 10^6 - 7.192\times 10^6 = -2.697\times10^6\,\mathrm{V}\) The electric potential at point \((4, 0)\) due to these charges is \(-2.697\times10^6\,\mathrm{V}\).