Simple Nature - Light and Matter

puter software:V = kλ ln(z + √ )∣ ∣∣z 2 + R 2 +L/2−L/2(L/2 + √ )L= kλ ln2 /4 + R 2−L/2 + √ L 2 /4 + R 2The expression inside the parentheses can be simplified a little.Leaving out some tedious algebra, the result is( √ )LV = 2kλ ln2R + 1 + L24R 2This can readily be differentiated to find the field:E R = − dVdR= (−2kλ) −L/2R2 + (1/2)(1 + L 2 /4R 2 ) −1/2 (−L 2 /2R 3 )L/2R + (1 + L 2 /4R 2 ) 1/2 ,or, after some simplification,kλLE R =R 2√ 1 + L 2 /4R 2For large values of R, the square root approaches one, and wehave simply E R ≈ kλL/R 2 = kQ/R 2 . In other words, the fieldvery far away is the same regardless of whether the charge is apoint charge or some other shape like a rod. This is intuitivelyappealing, and doing this kind of check also helps to reassureone that the final result is correct.The preceding example, although it involved some messy algebra,required only straightforward calculus, and no vector operationsat all, because we only had to integrate a scalar function to find thevoltage. The next example is one in which we can integrate eitherthe field or the voltage without too much complication.On-axis field of a ring of charge example 11⊲ Find the voltage and field along the axis of a uniformly chargedring.⊲ Integrating the voltage is straightforward.∫ k dqV =r∫dq= k √b 2 +∫z 2k= √ dqb 2 + z 2=kQ√b 2 + z 2 ,c / Example 11.Section 10.3 Fields by Superposition 575