/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 A charge of \(+2.0 \mathrm{mC}\)... [FREE SOLUTION] | 91影视

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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

A positively charged oil drop is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced $16 \mathrm{cm}$ apart. If the electric force on the drop is found to be $9.6 \times 10^{-16} \mathrm{N}\( and the potential difference between the plates is \)480 \mathrm{V}$ what is the magnitude of the charge on the drop in terms of the elementary charge \(e\) ? Ignore the small buoyant force on the drop.
A large parallel plate capacitor has plate separation of \(1.00 \mathrm{cm}\) and plate area of \(314 \mathrm{cm}^{2} .\) The capacitor is connected across a voltage of \(20.0 \mathrm{V}\) and has air between the plates. How much work is done on the capacitor as the plate separation is increased to $2.00 \mathrm{cm} ?$
A parallel plate capacitor is connected to a battery. The space between the plates is filled with air. The electric field strength between the plates is \(20.0 \mathrm{V} / \mathrm{m} .\) Then, with the battery still connected, a slab of dielectric \((\kappa=4.0)\) is inserted between the plates. The thickness of the dielectric is half the distance between the plates. Find the electric field inside the dielectric.
A cell membrane has a surface area of \(1.0 \times 10^{-7} \mathrm{m}^{2},\) a dielectric constant of \(5.2,\) and a thickness of \(7.5 \mathrm{nm}\) The membrane acts like the dielectric in a parallel plate capacitor; a layer of positive ions on the outer surface and a layer of negative ions on the inner surface act as the capacitor plates. The potential difference between the "plates" is \(90.0 \mathrm{mV}\). (a) How much energy is stored in this capacitor? (b) How many positive ions are there on the outside of the membrane? Assume that all the ions are singly charged (charge +e).
A van de Graaff generator has a metal sphere of radius \(15 \mathrm{cm} .\) To what potential can it be charged before the electric field at its surface exceeds \(3.0 \times 10^{6} \mathrm{N} / \mathrm{C}\) (which is sufficient to break down dry air and initiate a spark)?
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