Please answer this question clearly and step by step with the needed diagrams, a

Question

Please answer this question clearly and step by step with the needed diagrams, and With typed answers, not hand written.

2. An electron (labeled q1, with mass 9.11 x 10-31 kg) is 1.50 Hm to the left of a 2.00 nC charge (labeled q2, with mass 3.00 g). An infinite plane with a surface charge density of 4.00 nC/m2 lies 5.00 cm directly above the two charges. 3.00 cm above the plane and parallel to the plane is an infinite line of charge with a linear charge density of -3.50 nC/m. All of the charges and charge distributions are held fixed in place, and then q2 is released. What will be the initial acceleration (both magnitude and direction) of q2? (the figure below is not to scale) 3.00 cm 5.00 cm 1.50 nm

Explanation / Answer

Electric field at location of q2, E = sigma/2eo + 2klambda/r + kq1/d^2

= -(4e-9/[2*8.85e-12] + 2*9e9*-3.5e-9/0.08 ) j - 9e9*1.6e-19/1.5e-6^2 i

= 561.5j - 640 i

magnitude E = sqrt(561.5^2+640^2) = 851.4

acceleration a = q2*E/m = 2e-9*851.4/0.003 = 0.0005676 m/s^2

direction = arctan(561.5/640) = 41.26 degree below horizontal

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---