Solve using this steps please Instructions: Solve the problems below using the R

Question

Solve using this steps please Instructions: Solve the problems below using the RIDEA framework. Put all of your work in the exam book. The complete solution of each problem is worth 27 points. Problem 1 Figure above shows a thin, uniformly charged disk of radius R. If the disk carries surface charge density ?, using the superposition principle, one can show that the electric field at point P, a distance x from the disk along its axis has magnitude given by the following expression: Based on this result solve the following problem: The electric field at center of a uniformly charged disk has magnitude 3.0 x 10 N/C and the electric field at an axial point a distance x 12 cm from the disk is 6.3 kN/C. Find (a) the disk's radius (in cm) and (b) its charge density (in ?0hn

Explanation / Answer

for the given question

electric field on the axis of the charged disc is

E(z) = 2*pi*k*sigma(1 - z/sqrt(z^2 + R^2)) where sigma is the uniform charge density on the disc

hence

1. charge density sigma

coloumbs constant = k

radius of disc = R

distnace form the center of the dic on it saxis = z

now,

E(z = 0) = 3*10^5

hence

2*pi*k*sigma(1 - 0) = 3*10^5

hence

2*pi*k*sigma = 3*10^5

hence

E = 3*10^5(1 - z/sqrt(z^2 + R^2))

also, for z = 0.12 m

E = 6.3*10^3 N/C

hence

6.3*10^3 = 3*10^5(1 - 0.12/sqrt(0.12^2 + R^2))

R = 0.02498797706810 m = 2.498797706810 cm

hence

E = 3*10^5(1 - z/sqrt(6.2439899*10^-4 + z^2))

2. for deriving the electric field on the axis of a disc, we first observe that it is the superposition of electric fields due to all the concentric rings on the uniformly charged disc at a point on the axis of these rings. now, we know this expression already for radius r, so we integrate individual fields through radius R to get the final expression of electric field

3. to solve the problem, we have unknowsn like sigma and R

but we have value of E(z) for two values of z

hence we have two unknowns and two equations

so we find these two unknowns, sigma and R

and then we plug them into equation of electric field to find electric field at any point on the axis of the disc

4. hence the evaluation of the answer gives us

E = 3*10^5(1 - z/sqrt(6.2439899*10^-4 + z^2))

5. accoring to the answer

E = 3*10^5(1 - 1/sqrt(6.2439899*10^-4/z^2 + 1))

for z-> inf

E = 0 ( which is true)

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