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Chapter 11: Problem 1585

Electric charges of \(+10 \mu \mathrm{c}, 5 \mu \mathrm{c},-3 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed at the corners of a square of side $\sqrt{2 m}\( the potential at the centre of the square is \)\ldots \ldots$ (A) \(1.8 \mathrm{~V}\) (B) \(1.8 \times 10^{5} \mathrm{~V}\) (C) \(1.8 \times 10^{6} \mathrm{~V}\) (D) \(1.8 \times 10^{4} \mathrm{~V}\)

Short Answer

Expert verified
The potential at the center of the square is (D) \(1.8 \times 10^{4} V\).

Step by step solution

01

Determine the distance to the center

Since the charges are on a square with side length \(\sqrt{2m}\), we need to find the distance of each charge to the center of the square. Since the square is symmetric, the distances for all the charges are the same. We can find the distance from one corner to the center by using the Pythagorean theorem: \(r = \sqrt{(\frac{\sqrt{2m}}{2})^2 + (\frac{\sqrt{2m}}{2})^2} = \sqrt{\frac{2m}{4}+\frac{2m}{4}} = \sqrt{\frac{2m}{2}} = \sqrt{m}\)
02

Calculate potential for each charge

Now we need to calculate the electric potential due to each charge at the center of the square. We will use the formula for electric potential, \(V = \frac{kQ}{r}\), and the distances calculated in step 1. Charge 1: \(V_1 = \frac{9 \times 10^9 (+10 \times 10^{-6})}{\sqrt{2}} \) Charge 2: \(V_2 = \frac{9 \times 10^9 (+5 \times 10^{-6})}{\sqrt{2}} \) Charge 3: \(V_3 = \frac{9 \times 10^9 (-3 \times 10^{-6})}{\sqrt{2}} \) Charge 4: \(V_4 = \frac{9 \times 10^9 (+8 \times 10^{-6})}{\sqrt{2}} \)
03

Sum the potentials

Since the electric potential is a scalar quantity, we can simply add the potentials from each charge to find the net electric potential at the center of the square. \(V_\text{net} = V_1 + V_2 + V_3 + V_4\) Substituting the expressions from step 2: \(V_\text{net} = \frac{9 \times 10^9}{\sqrt{2}}(10 \times 10^{-6} + 5 \times 10^{-6} - 3 \times 10^{-6} + 8 \times 10^{-6})\)
04

Simplify and find the answer

Now, let's simplify and calculate the net electric potential: \(V_\text{net} = \frac{9 \times 10^9}{\sqrt{2}}(20 \times 10^{-6})\) \(V_\text{net} = 1.8 \times 10^4 V\) So, the correct answer is (D) \(1.8 \times 10^{4} V\).

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Most popular questions from this chapter

Two air capacitors \(A=1 \mu F, B=4 \mu F\) are connected in series with $35 \mathrm{~V}\( source. When a medium of dielectric constant \)\mathrm{K}=3$ is introduced between the plates of \(\mathrm{A}\), change on the capacitor changes by (A) \(16 \mu \mathrm{c}\) (B) \(32 \mu \mathrm{c}\) (C) \(28 \mu \mathrm{c}\) (D) \(60 \mu \mathrm{c}\)

There are 10 condensers each of capacity \(5 \mu \mathrm{F}\). The ratio between maximum and minimum capacities obtained from these condensers will be (A) \(40: 1\) (B) \(25: 5\) (C) \(60: 3\) (D) \(100: 1\)

Two point charges repel each other with a force of \(100 \mathrm{~N}\). One of the charges is increased by \(10 \%\) and other is reduced by \(10 \%\). The new force of repulsion at the same distance would be \(\ldots \ldots \mathrm{N}\). (A) 121 (B) 100 (C) 99 (D) 89

Electric potential at any point is $\mathrm{V}=-5 \mathrm{x}+3 \mathrm{y}+\sqrt{(15 \mathrm{z})}$, then the magnitude of the electric field is \(\ldots \ldots \ldots \mathrm{N} / \mathrm{C}\). (A) \(3 \sqrt{2}\) (B) \(4 \sqrt{2}\) (C) 7 (D) \(5 \sqrt{2}\)

Capacitance of a parallel plate capacitor becomes \((4 / 3)\) times its original value if a dielectric slab of thickness \(t=d / 2\) is inserted between the plates (d is the separation between the plates). The dielectric constant of the slab is (A) 8 (B) 4 (C) 6 (D) 2
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