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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 parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) and is connected to a \(12-\mathrm{V}\) battery. (a) What is the magnitude of the charge on each plate? (b) If the plate separation is doubled while the plates remain connected to the battery, what happens to the charge on each plate and the electric field between the plates?
It has only been fairly recently that 1.0 -F capacitors have been readily available. A typical 1.0 -F capacitor can withstand up to \(5.00 \mathrm{V}\). To get an idea why it isn't easy to make a 1.0 -F capacitor, imagine making a \(1.0-\mathrm{F}\) parallel plate capacitor using titanium dioxide \((\kappa=90.0\) breakdown strength \(4.00 \mathrm{kV} / \mathrm{mm}\) ) as the dielectric. (a) Find the minimum thickness of the titanium dioxide such that the capacitor can withstand \(5.00 \mathrm{V}\). (b) Find the area of the plates so that the capacitance is \(1.0 \mathrm{F}\)
A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) There is a charge of \(0.80 \mu \mathrm{C}\) on each plate. How much work must be done by an external agent to double the plate separation while keeping the charge constant?
Draw some electric field lines and a few equipotential surfaces outside a positively charged conducting cylinder. What shape are the equipotential surfaces?
A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\) The parallel plates are separated by \(0.40 \mathrm{mm}\) of bakelite. What is the capacitance of this capacitor?
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