PHY2054 Discussion-Fall 08

September 12, 2008 PHY2054 Discussion-Fall ‘08 Quiz 2 (Chapter 15.7-16.3)

Name: UFID:

*1. (5pts) The dome of a Van de Graaff generator receives a charge of 3.00×10-4 C. Find the strength of the electric field at the outer surface of the dome, assuming the dome is a conducting spherical shell with a radius of 1.50 m.

We apply Gauss’s law to a spherical Gaussian surface of radius 1.5 m. Since the entire charge is inside the Gaussian surface, we have

9 -4 6 E(4πr²) = Q/ε0 ⇒ E = Q/(4πε0r²) = kQ/r² = 8.99×10 ×3×10 /1.5² = 1.20×10 N/C

*2. (5pts) An electric field with an intensity of 5.00 kN/C is applied along the x-axis. What is the electric flux through a rectangular plane 0.400 m wide and 0.600 long if the plane contains y-axis and its normal makes an angle of 60.0˚ with the x-axis.

Using the definition of electric flux, we have Φ = EAcosθ = 5×103×(0.4×0.6)×cos60˚ = 6.00×102 N m²/C ***3. (5pts) A 2.00-kg block is attached to the lower end of a spring lying along a 30.0˚ incline. The coefficient of kinetic friction between the incline and the block is 0.25. The spring has a force constant of 150 N/m and it is fixed at the upper end. The block is charged and the system is immersed in a uniform electric field with an intensity of 6.00×105 N/C directed vertically downward. The block is released from rest when the spring is unstretched. If the block slides 0.200 m down the incline before it turns around for the first time, what is the charge given to the block?

We take +x-axis along the incline and +y perpendicular to the incline. Since the block does not move in the y direction, the normal force is expressed as

ΣFy = 0 ⇒ n-mgcosθ-qEcosθ = 0 ⇒ n = (mg+qE)cosθ Applying work-energy theorem to the system, we have

Wnc = ΔE ⇒ -μ(mg+qE)cosθΔx = (1/2)kΔx²-mgsinθΔx-qEsinθΔx ⇒ qE(sinθ-μcosθ) = (1/2)kΔx-mg(sinθ-μcosθ) ⇒ q = kΔx/[2E(sinθ-μcosθ)]-mg/E = 150×0.2/[2×6×105(sin30˚-0.25cos30˚)]-2×9.8/(6×105) = 5.55×10-5 C

**4. (5pts) Three charges q1 =2.00 μC, q2 = -4.00 μC and q3 = 6.00 μC are placed at three vertices of a rectangle, q1 at the upper left corner, q2 at the lower left and q3 at the lower right. q1 and q2 are separated by 0.300 m and q2 and q3 are separated by 0.400 m. What is the work required to move a -8.00-μC charge from infinity to the fourth vertex?

The length of the diagonal is r2 = √r1²+r3²) = √(0.3²+0.4²) = 0.5 m The required work is work done by an external force equal and opposite to the electric force. Thus we have,

Wex = -Wel = ΔPE = qΔV = qk(q1/r1+q2/r2+q3/r3) = -8×10-6×8.99×109×(2/0.4-4/0.5+6/0.3)×10-6 = -1.22 J