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Chapter 11: Problem 1573

Three charges \(2 q,-q,-q\) are located at the vertices of an equilateral triangle. At the centre of the triangle. (A) The Field is Zero but Potential is non - zero (B) The Field is non - Zero but Potential is zero (C) Both field and Potential are Zero (D) Both field and Potential are non - Zero

Short Answer

Expert verified
The electric field and potential at the center of the equilateral triangle are both non-zero. The correct statement is (D) Both field and Potential are non-Zero.

Step by step solution

01

Calculate Individual Electric Field Vectors

For calculating the electric field at the center of the equilateral triangle, we need to consider the individual electric field from each charge on the center. For an equilateral triangle, each angle = \(120^\circ\). Let's refer to vertices of the triangle as A, B, and C, with charges \(2q\), \(-q\), and \(-q\) respectively, and the center of the triangle as O. Divide the problem into three parts: 1. Electric field at O due to the charge at A (positive charge) 2. Electric field at O due to the charge at B (negative charge) 3. Electric field at O due to the charge at C (negative charge) Calculate each electric field vector separately.
02

Add Electric Field Vectors

Now, add the individual electric field vectors at point O. Since the electric field vectors due to the negative charges at vertices B and C are equal in magnitude but opposite in direction, their sum will have the same magnitude but will be directed in the opposite direction to the positive charge at vertex A. The net electric field at point O comes from the difference between the electric field due to the positive charge and the sum of the electric fields due to the negative charges.
03

Calculate Individual Electric Potentials

The next step is to calculate the electric potential at point O. Calculate the electric potentials at O due to each of the charges at A, B and C. Equation for electric potential V is given by: \(V = \frac{k*q}{r}\) where V is electric potential, k is Coulomb's constant, q is the charge, and r is the distance from the charge to the point.
04

Add Electric Potentials

Now, add the individual electric potentials at point O. Since electric potential is a scalar, we can directly add potentials due to all three charges: \(V_{total} = V_A + V_B + V_C\)
05

Determine the Correct Statement

Based on the calculations in the previous steps, we can determine which statement is correct about the electric field and potential at the center of the equilateral triangle. (A) If the Field is Zero but the Potential is non-zero; (B) If the Field is non-Zero but the Potential is zero; (C) If both the Field and Potential are Zero; (D) If both the Field and Potential are non-Zero. From our calculations, we find that the electric field at point O is non-zero, and the electric potential is also non-zero. Therefore, the correct statement is: (D) Both field and Potential are non-Zero.

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Most popular questions from this chapter

The inward and outward electric flux for a closed surface in units of \(\mathrm{Nm}^{2} / \mathrm{C}\) are respectively \(8 \times 10^{3}\) and $4 \times 10^{3}\(. Then the total charge inside the surface is \)\ldots \ldots \ldots \ldots . . \mathrm{c}$. (A) \(\left[\left(-4 \times 10^{3}\right) / \epsilon_{0}\right]\) (B) \(-4 \times 10^{3}\) (C) \(4 \times 10^{3}\) (D) \(-4 \times 10^{3} \mathrm{E}_{0}\)

The potential at a point \(\mathrm{x}\) (measured in \(\mu \mathrm{m}\) ) due to some charges situated on the \(\mathrm{x}\) -axis is given by \(\mathrm{V}(\mathrm{x})=\left[(20) /\left(\mathrm{x}^{2}-4\right)\right]\) Volt. The electric field at \(\mathrm{x}=4 \mu \mathrm{m}\) is given by (A) \((5 / 3) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (B) \((10 / 9) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (C) \((10 / 9) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (D) \((5 / 3) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction

Two positive point charges of \(12 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed \(10 \mathrm{~cm}\) apart in air. The work done to bring them $4 \mathrm{~cm}$ closer is (A) Zero (B) \(4.8 \mathrm{~J}\) (C) \(3.5 \mathrm{~J}\) (D) \(-5.8 \mathrm{~J}\)

A small sphere whose mass is \(0.1 \mathrm{gm}\) carries a charge of $3 \times 10^{-10} \mathrm{C}\( and is tie up to one end of a silk fiber \)5 \mathrm{~cm}$ long. The other end of the fiber is attached to a large vertical conducting plate which has a surface charge of \(25 \times 10^{-6} \mathrm{Cm}^{-2}\), on each side. When system is freely hanging the angle fiber makes with vertical is \(\ldots \ldots \ldots\) (A) \(41.8^{\circ}\) (B) \(45^{\circ}\) (C) \(40.2^{\circ}\) (D) \(45.8^{\circ}\)

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
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