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Chapter 20: Problem 53

Two identical small metal spheres initially carry charges \(q_{1}\) and \(q_{2} .\) When they're \(1.0 \mathrm{m}\) apart, they experience a \(2.5-\mathrm{N}\) attractive force. Then they're brought together so charge moves from one to the other until they have the same net charge. They're again placed \(1.0 \mathrm{m}\) apart, and now they repel with a \(2.5-\mathrm{N}\) force. What were the original charges \(q_{1}\) and \(q_{2} ?\)

Short Answer

Expert verified
The original charges were positive and negative of the same value that will be computed from \(X\).

Step by step solution

01

Computing the absolute product of the charges

The absolute value of the product of the charges is given by multiplying the force by the square of the distance, and dividing by Coulomb's constant. Let's denote this quantity by \(X\), so \(X = \frac{F \cdot r^2}{k} = \frac{2.5 \times (1.0)^2}{8.99 \times 10^9}\). Calculate the value of \(X\).
02

Examine the conditions

Note that the charges were initially opposite, and they ended up with the same sign. If their values are \(q1\) and \(q2\) respectively, then \(q1 = - q2\) initially, which means their product is \( -X\). After redistribution of charges, \(q1 = q2\) which means their product is \(X\). This implies that their new common value is the square root of \(X\). Calculate the value of \(q1\) and \(q2\).
03

Find the original charges

Using the relations identified in the previous step, the original charges can be found. Since \(q1 = -q2\) initially, the original charges were \(q1\) and \(-q1\), where \(q1\) is the value found in step 2.

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Most popular questions from this chapter

You're \(1.5 \mathrm{m}\) from a charge distribution whose size is much less than \(1 \mathrm{m}\). You measure an electric field strength of \(282 \mathrm{N} / \mathrm{C}\) You move to a distance of \(2.0 \mathrm{m},\) and the field strength becomes 119 N/C. What's the net charge of the distribution? (Hint: Don't try to calculate the charge. Determine instead how the field decreases with distance, and from that infer the charge.)

You're taking physical chemistry, and your professor is discussing molecular dipole moments. Water, he says, has a dipole moment of "1.85 debyes," while carbon monoxide's dipole moment is only "0.12 debye." Your physics professor wants these moments expressed in SI. She tells you that the atomic separation in these two covalent compounds is about the same, and asks what that indicates about the way shared charge is distributed. What do you tell her?

Find the magnitude of the electric force on a \(2.0-\mu \mathrm{C}\) charge in a \(100-\mathrm{N} / \mathrm{C}\) electric field.

An electron at Earth's surface experiences a gravitational force \(m_{c} g .\) How far away can a proton be and still produce the same force on the electron? (Your answer should show why gravity is unimportant on the molecular scale!)

A dipole lies on the \(y\) -axis and consists of an electron at \(y=0.60 \mathrm{nm}\) and a proton at \(y=-0.60 \mathrm{nm} .\) Find the electric field (a) midway between the two charges; (b) at the point \(x=2.0 \mathrm{nm}\) \(y=0 \mathrm{nm} ;\) and \((\mathrm{c})\) at the point \(x=-20 \mathrm{nm}, y=0 \mathrm{nm}\)
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