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Chapter 20: Problem 51

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.)

Short Answer

Expert verified
Using the known parameters and the inverse-square dependence of the electric field, the net charge of the distribution can be found: \(Q = E1 \cdot r1^2 / k\).

Step by step solution

01

Identify known quantities from the problem

From the problem, we have two electric field measurements taken at two different distances: E1 = 282 N/C at r1 = 1.5 m, and E2 = 119 N/C at r2 = 2.0 m.
02

Determine how the field decreases with distance

We find the ratio of the field strengths and distances: \((E2/E1) = (r1/r2)^2\). From this, we infer that the field follows an inverse-square law behavior, which is consistent with the electric field around a point charge.
03

Calculate the net charge

Since the system follows an inverse-square law, the charge Q can be calculated from one position using the formula \(E1 = kQ/r1^2\). Solving for Q, we find that \(Q = E1 \cdot r1^2 / k\). Plugging the known values of E1, r1, and Coulomb's constant (k = 8.99 x 10^9 N m2/C2), we find the net charge.

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Most popular questions from this chapter

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