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Chapter 17: Problem 20

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

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

Step by step solution

01

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: \(r_2=\sqrt{(4-0)^2+(0-3)^2} = 5\,\mathrm{m}\)
02

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 \(Q_2\): \(V_2 = \frac{(8.99\times10^9)(-4.0\times10^{-3})}{5} = -7.192\times10^6\,\mathrm{V}\)
03

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 \(-2.697\times10^6\,\mathrm{V}\).

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Most popular questions from this chapter

Before a lightning strike can occur, the breakdown limit for damp air must be reached. If this occurs for an electric field of $3.33 \times 10^{5} \mathrm{V} / \mathrm{m},$ what is the maximum possible height above the Earth for the bottom of a thundercloud, which is at a potential $1.00 \times 10^{8} \mathrm{V}$ below Earth's surface potential, if there is to be a lightning strike?
Capacitors are used in many applications where you need to supply a short burst of energy. A \(100.0-\mu \mathrm{F}\) capacitor in an electronic flash lamp supplies an average power of \(10.0 \mathrm{kW}\) to the lamp for $2.0 \mathrm{ms}$. (a) To what potential difference must the capacitor initially be charged? (b) What is its initial charge?
In \(1911,\) Ernest Rutherford discovered the nucleus of the atom by observing the scattering of helium nuclei from gold nuclei. If a helium nucleus with a mass of \(6.68 \times 10^{-27} \mathrm{kg},\) a charge of \(+2 e,\) and an initial velocity of \(1.50 \times 10^{7} \mathrm{m} / \mathrm{s}\) is projected head-on toward a gold nucleus with a charge of \(+79 e\), how close will the helium atom come to the gold nucleus before it stops and turns around? (Assume the gold nucleus is held in place by other gold atoms and does not move.)
A spherical conductor of radius \(R\) carries a total charge Q. (a) Show that the magnitude of the electric field just outside the sphere is \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the charge per unit area on the conductor's surface. (b) Construct an argument to show why the electric field at a point \(P\) just outside any conductor in electrostatic equilibrium has magnitude \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the local surface charge density. [Hint: Consider a tiny area of an arbitrary conductor and compare it to an area of the same size on a spherical conductor with the same charge density. Think about the number of field lines starting or ending on the two areas.]
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