In the figure a nonconducting rod of length L = 8.30 cm has charge -q = -4.55 fC

Question

In the figure a nonconducting rod of length L = 8.30 cm has charge -q = -4.55 fC uniformly distributed along its length. (a) What is the linear charge density of the rod? What are the (b) magnitude and (c) direction (positive angle relative to the positive direction of the x axis) of the electric field produced at point P, at distance a = 13.3 cm from the rod? What is the electric field magnitude produced at distance a = 56 m by (d) the rod and (e) a particle of charge -q = -4.55 fC that replaces the rod?

Explanation / Answer

capacitance C0 = eo*A/d

energy stored U0 = Q^2/(2C0)

after the battery is disconnected

charge Qf = Q

Capacitance C = eo*A/(9.7*d) = C0/9.7

energy stored Uf = Q^2/(2*C) = Q^2/(2*C0/9.7) = 9.7*Q^2/(2*C0)

Uf = 9.7*U0

Uf/U0 = 9.7 = 10

( a)

Linear charge density Lambda = q/L = -4.55*10^-15/(8.3*10^-2) = -54.8 fC/m

(b)

consider a small length dx of charge dq located a distance x from p

dq = lambda*dx

lambda = charge density

electric field due to dq , dEx = k*dq/x^2

Ex = integration dEx

Ex = integration k*lambda*dx/x^2 from x = a to x = L+a

Ex = k*lambda*(-1/x) from x = a to x = L+a

Ex = k*lambda*(1/a - 1/(L+a))

Ex = -9*10^9*54.8*10^-15*(1/0.133 - 1/(0.083+0.133))

Ex = -0.001424 N/C

magnitude = 0.001424 N/C

direction = 180 degrees from positive x axis

----------------

(d)

E = -9*10^9*54.8*10^-15*(1/56 - 1/(0.083+56))

E = 1.3*10^-8 N/C

(e)

for a particle E = k*q/(L+a)^2

E = 9*10^9*4.55*10^-15*/(56+0.083)^2 = 1.3*10^-8 N/C

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