As part of the optics in a prototype scanning electron microscope, you have a un
ID: 2223535 • Letter: A
Question
As part of the optics in a prototype scanning electron microscope, you have a uniform circular ring of charge Q=4.80 microCoulombs and radius R=1.30 cm located in the x-y plane, centered on the origin as shown in the figure. If z is much smaller than R then E is proportional to z. (You should verify this by taking the limit of your expression for E for z much smaller than R.) If you place an electron on the z-axis near the origin it experiences a force Fz=-cz, where c is a constant. Obtain a numerical value for c. (To think about: what kind of motion would that electron undergo if you let it go?)Explanation / Answer
elementary charge on the ring dq= R*dt, where =Q/(2piR) is linear density, dt is elementary polar angle around z-axis, R*dt is elementary length on the ring
strength of electric field for point P=(0,0,z) by charge dq is vector
dE=k*dq *r/|r|^3, where vector r=(R*cost, R*sint, z), distance |r| =(R^2 +z^2), constant k=8.988·10^9 SI units;
net E for point P is dE for t=0 until 2pi, i.e.
E= k*dq *r/|r|^3 = (k R / |r|^3) dt*r;
since integration for t=0 until 2pi will compensate x and y components of vector r we may discard these component in integration, so that net strength now looks:
E = (k R z / |r|^3) dt = (k R z / |r|^3) *2pi = kQz / (R^2 +z^2)^3
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