(8c24p25) The figure on the left shows a positively charged plastic rod of lengt

Question

(8c24p25) The figure on the left shows a positively charged plastic rod of length L = 2.00 m and uniform linear charge density 1.00?10-3 C/m. Setting V= 0 at infinity and considering Fig. 25-13 and Eq 25-35, find the electric potential at point P for d = 1.040 m. The figure on the right shows an identical rod, except that it is split in half and the right half is negatively charged; the left and right halves have the same magnitude ? of uniform linear charge density, what is the electric potential at point P in the figure?

Explanation / Answer

a)

l = L/2 = 2/2 = 1 m

V = 2k * ln((l+(l2+a2))/(a))

V = 2*9e9*1e-3 * ln((1+(1.040*1.040+1*1))/1.040)

V = 2*9e9*1e-3 * ln((1+1.44278)/1.040)

V = 1.54e7 V

b)

V = 0

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.