A conducting plate with a uniform surface charge density s = +5.5 x 10 -8 C/m 2

Question

A conducting plate with a uniform surface charge densitys = +5.5 x 10-8C/m2 is spaced a distance d = 2.1 cm above andparallel to an identical plate with uniform surface charge density-s. The spacing d issmall compared to the x- and z-dimensions of theplates.

(a) What is the y-component of the electric fieldmidway between the plates and far from the edges of the plates?

Ey = V/m     *
-6.21e3   OK

HELP:  Treat the plates as infinitein the x- and z-directions.

(b) What is the electric potential of the top plate relative tothe bottom plate?

V1 - V2 = V     *
130.41   OK

HELP:  Recall the basic definition ofthe electric potential in terms of electric field.

(c) A particle of mass m = 5.4 g and charge Q= -4.9 x 10-3 C rests at the bottom side of the topplate. How much work would have to be done on the particle to bringit to the bottom plate? (Ignore gravity.)

W = J    

(d) If the same particle is subsequently released at rest fromthe bottom plate, with what speed will it impact the top plate?

v = m/s    

(e) Suppose instead the upper plate was moved closer to thelower plate, reducing the spacing to d' = 1.2 cm, withoutchanging the net charges on the upper or lower plates. What wouldthe impact speed be if the same particle were released at rest fromthe lower plate in this case?

v' = m/s    

Explanation / Answer

E = 6.21*103 V/m, V1 - V2 = 130.4V, m = 5.4 g, Q = -4.9 x 10-3 C c) W = increase of the energy = QV2 - QV1 =-Q(V1 - V2) = 0.639 J d) QV2 = QV1 + mv2/2 v = [-2Q(V1 - V2)/m] =(2W/m) = 15.4 m/s e) V1' - V2' = E*d' = (V1 -V2)d'/d v' = [-2Q(V1' - V2')/m] =[-2Q(V1 - V2)d'/(dm)] = v(d'/d)= 15.4*(1.2/2.1) = 11.6 m/s

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