I only need D ---> F (a) as a single charged particle at x = 2.00 m 41.4 V (b) a
ID: 1268262 • Letter: I
Question
I only need D ---> F
(a) as a single charged particle at x = 2.00 m
41.4 V
(b) as two 4.60 nC charged particles at x = 1.50 m and x = 2.50 m
44.16 V
(c) as three 3.070 nC charged particles at x = 1.33 m, x = 2.00 m, and x = 2.67 m.
44.94 V
(d) Now do the necessary integral to get the exact result for the potential at the origin. You should first do the integral symbolically before substituting the necessary values.
V
(e) Use your exact expression to find the potential at these additional locations.
at x = ?1 m V
at x = ?2 m V
(f) Take the necessary derivative to find an expression for the electric field strength at any location to the left of the left end of the filament. Use this expression to find the electric field strength at the locationx = ?1 m.
V/m
Explanation / Answer
help from
A uniformly charged filament of length lies along the x axis between x = a = 1.00 m and x = a + = 3.00 m. The total charge on the filament is 2.10 nC. Calculate successive approximations for the electric potential at the origin by modeling the filament in the following ways.
(a) as a single charged particle at x = 2.00 m
(b) as two 1.05 nC charged particles at x = 1.5 m and x = 2.5 m
(c) as four 0.525 nC charged particles at x = 1.25 m, x = 1.75 m, x = 2.25 m, and x = 2.75 m.
(d) Determine the exact potential at the origin correct to the third decimal place
answer
.a) V = k*q/r = 9.0x10^9*2.10x10^-9/2.00 = 9.45V
b) V = k*q1/r1 + k*q2/r2 = 9.0x01^9*(1.05x01^-9/1.5 + 1.05x10^-9/2.5 ) = 10.08V
c) V = k*q1/r1 + k*q2/r2 + k*q3/r3 + k*q4/r4 =
9.0x10^9*0.525x10^-9*(1/1.25 + 1/1.75 + 1/2.25 + 1/2.75) = 10.298V
d) Here we have dV = k*Q/L*dx/(x)
Now integrate from x = 1.00 to 3.00
So V = k*Q/L*ln(x) from 1 to 3 = k*Q/L*(ln(3) - ln(1)) = k*Q/L*ln(3)
= 9.0x10^9*2.10x10^-9/2.0m*ln(3) = 10.382V