Question

#8 plz

Your session has expired. Lab Req x 1 , Google Translate | ㅁ ElectricCharge and Elec: X % University PhysicstHon x -G consider Ar x 10 ริเ Need Help? Rase the axis of the rod at a point 30.0 cm from its center A rod 10.0 cm long is uniformly charged and has a total charge of -20.0 uC. Determine the magnitude and direction of the electric field along N/C direction nhat is the electric field Cin MW/C) dve to the charged disk at s (a) a-5.00 m MN/C MacBook Air

Answer 1

linear density = charge per unit length, lambda = Q/L

consider a small length dx at a point x on rod

dq = lambda*dx

dE = k*dq/x^2 = k*lambda*dx/x^2

E = integration dE from d- L/2 to d + L/2

E = k*lambda integration dx/x^2

E = k*Q/L*( - 1/x) from x = d - L/2 to x = d + L/2

E = k*Q/L*( 1/(d - L/2) - 1/ ( d + L/2) )

d = 30 cm = 0.3 m

L = 10 cm = 0.1 m

E = 9*10^9*20*10^-6/0.1*( 1/ ( 0.3 - 0.05) - 1/(0.3 + 0.05) )

E = 2.06*10^6 N/C

direction towards the rod

$\fn_cm \\Electric\;field\;due\;to\;charged\;disk\\\\ E=k\sigma 2\pi \left [1-\frac{z}{\sqrt{z^2+R^2}} \right ]\\\\ (a)\\E=9\times 10^9\times 6.7\times 10^{-3}\times 2\times \pi \left [1-\frac{0.05}{\sqrt{0.05^2+0.4^2}} \right ]=332MN/C\\\\ (b)\\E=9\times 10^9\times 6.7\times 10^{-3}\times 2\times \pi \left [1-\frac{0.1}{\sqrt{0.1^2+0.4^2}} \right ]=287MN/C\\\\$