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Chapter 13: Problem 62

Coulomb's law is applicable to, (A) Point charges (B) Spherical charges (C) Like charges (D) All of these

Short Answer

Expert verified
The correct answer is (D) All of these, as Coulomb's law is applicable to point charges, spherical charges, and like charges.

Step by step solution

01

Defining Coulomb's Law

Coulomb's law describes the electrostatic force of attraction or repulsion between two charged particles. The force between two point charges is given by the formula: \(F = k \cdot \frac{q_1 \cdot q_2}{r^2}\) where: - F is the electrostatic force, - k is the electrostatic constant (鈮� 8.988 x 10^9 Nm^2/C^2), - q1 and q2 are the magnitudes of the point charges, - r is the distance between the point charges. Now, let's evaluate each option:
02

(A) Point charges

Point charges are charges with zero dimensions, which means that their charge is concentrated at a single point in space. As Coulomb's law formula involves the charges q1 and q2 and the distance r between these charges, it is applicable to point charges.
03

(B) Spherical charges

Spherical charges are charges distributed on the surface of a sphere. When dealing with two spheres with uniformly distributed charges on their surfaces, we can consider the charges on each sphere to be concentrated at their center. Therefore, Coulomb's law is applicable to spherical charges.
04

(C) Like charges

Like charges are charges with the same sign, either both positive or both negative. The formula of Coulomb's law implies that like charges repel each other (positive x positive = positive, negative x negative = positive), while opposite charges attract each other (positive x negative = negative). Thus, Coulomb's law is applicable to like charges too.
05

(D) Conclusion

Since Coulomb's law is applicable to point charges, spherical charges, and like charges, the correct answer is (D) All of these.

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Most popular questions from this chapter

A capacitor of capacitance \(\mathrm{C}\) is fully charged by a \(200 \mathrm{~V}\) supply. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat \(2.5 \times 10^{2} \mathrm{~J} / \mathrm{kg}-\mathrm{K}\) and of mass \(0.1 \mathrm{~kg}\). If temperature of the block rises by \(0.4 \mathrm{~K}\). Find the value of \(C\).

Two point charges \(+8 q\) and \(-2 q\) are located at \(x=0\) and \(x=L\), respectively. The location of a point on the \(x\)-axis at which the net electric field due to these two point charges is zero (A) \(2 L\) (B) \(\frac{L}{4}\) (C) \(8 L\) (D) \(4 L\)

Three charges \(Q,+q\), and \(+q\) are placed at the vertices of a right angle triangle (isosceles triangle) as shown. The net electrostatic energy of the configuration is zero, if \(Q\) is equal to (A) \(\frac{-q}{1+\sqrt{2}}\) (B) \(\frac{-2 q}{2+\sqrt{2}}\) (C) \(-2 q\) (D) \(+q\)

If three moles of monatomic gas is mixed with 1 moles diatomic gas, the resultant value of \(\gamma\) for the mixture is (A) \(1.66\) (B) \(1.50\) (C) \(1.40\) (D) \(1.57\)

On moving a charge of \(20 \mathrm{C}\) by \(2 \mathrm{~cm}, 2 \mathrm{~J}\) of work is done. The potential difference between the points (A) \(0.1 \mathrm{~V}\) (B) \(8 \mathrm{~V}\) (C) \(2 \mathrm{~V}\) (D) \(0.5 \mathrm{~V}\)
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