PHY133 Test1 Review 1) Two infinite-plane nonconducting thin sheets of uniform s

Question

PHY133 Test1 Review 1) Two infinite-plane nonconducting thin sheets of uniform surface 13uC/m2 and -9uC/m2 are parallel to each other and 0.2 m apart. What are the electric fields between the two sheets and outside of them? 2) A positive charge of 18 nC is evenly distributed along a straight rod of length 4.0 m that is bent into a circular arc with a radius of 2.0 m. Find the magnitude of the electric field at the center of curvature of the arc? 3) Find the electric field at a point P due to a 10 nC charged disk of radius 2 cm. The charges are fixed and uniformly distributed on the disk. The point P is 40 cnm away from the disk. Now you drill a hole of radius 0.4cm at the center of the disk Find the electric field of this new disk with the hole at the center.

Explanation / Answer

electric field due to infinite sheet E = sigma/(2*epsilon_0)

1)

betwen the two sheets the two fields are in same direction

E = E1 + E2 = 13*10^-6/(2*8.85*10^-12) + 9*10^-6/(2*8.85*10^-12)

E = 1.243*10^6 N/C

outside the two sheets the two fields are in opposite direction

E = E1 - E2 = 13*10^-6/(2*8.85*10^-12) - 9*10^-6/(2*8.85*10^-12)

E = 0.226*10^6 N/C

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2)

angle subtended at the center theta = 4/2 = 2 radians = 114.6 degrees

charge density , lambda = Q/L = 18/4 = 4.5 nC

electric field at the center = 2*k*lambda*sin(theta/2)/r

E = 2*9*10^9*4.5*10^-9*sin(114.6/2)/2 = 34.08 N/C

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3)

electric field deu to charged disk E = k*sigma*2*pi*(1 - x/sqrt(x^2+R^2))

sigma = Q/(pi*R^2) = 10/(pi*0.02^2) nc/m^2 = 7958 nC/m^2

x = 40 cm = 0.4 m

E = 9*10^9*7958*10^-9*2*pi*(1-0.4/sqrt(0.4^2+0.02^2))

E = 561.5 N/C

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