problem23_90

University Physics with Modern Physics with Mastering Physics (11th Edition)

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Unformatted text preview: 23.90: For an infinitesimal slice of a finite cylinder, we have the potential: dV = k dQ ( x − z )2 + R 2 kQ L L 2 = kQ dz L ( x − z )2 + R 2 dz 2 2 ⇒V = ⇒V = −L 2 ∫ ( x − z) + R = kQ L L 2−x −L 2 − x ∫ du u + R2 2 where u = x − z. kQ ( L 2− x) 2 + R 2 + ( L 2 − x ) ln on the cylinder's axis. L ( L 2 + x)2 + R 2 − L 2 − x b) For L << R : V ≈ 2 2 2 2 kQ (L 2 − x) + R + L 2 − x kQ x − xL + R + L 2 − x ln ≈ ln L ( L 2 + x) 2 + R 2 − L 2 − x L x 2 + xL + R 2 − L 2 − x 2 2 2 2 kQ 1 − xL (R + x ) + ( L 2 − x) R + x ⇒V ≈ ln L 1 + xL ( R 2 + x 2 ) + (− L 2 − x) R 2 + x 2 2 2 2 2 kQ 1 − xL 2( R + x ) + ( L 2 − x) R + x ⇒V ≈ ln L 1 + xL 2( R 2 + x 2 ) + (− L 2 − x) R 2 + x 2 2 2 kQ 1 + L 2 R + x kQ L L ln 1 + ⇒V ≈ ln = − ln 1 − 2 2 2 2 2 R2 + x2 L L 1 − L 2 R + x 2 R + x kQ 2L kQ ⇒V ≈ = , which is the same as for a ring. L 2 x2 + R2 x2 + R2 c)  2 2 2 2 ∂V 2kQ ( L − 2 x) + 4 R − ( L + 2 x) + 4 R E=− = . ∂x ( L − 2 x) 2 + 4 R 2 ⋅ ( L + 2 x) 2 + 4 R 2 ( ) ...
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