The figure below shows three flat plates, all of very large area. The two thin p

Question

The figure below shows three flat plates, all of very large area. The two thin plates are made of insulating material and carry uniformly-distributed surface charge densities of ?a and ?b respectively. The thick metal plate has width w, and is initially uncharged.

1/ What is the magnitude of the electric field EA at the origin (the point marked A on the figure)? ANSWER (   |EA| = 2.8

The figure below shows three flat plates, all of very large area. The two thin plates are made of insulating material and carry uniformly-distributed surface charge densities of ?a and ?b respectively. The thick metal plate has width w, and is initially uncharged. What is the magnitude of the electric field EA at the origin (the point marked A on the figure)? ANSWER (|EA| = 2.8 times 105 N/C) What is the sign of the surface charge density ?LHS on the left-hand side of the thick metal plate? ANSWER (?LHS > 0) Compare the electric potential Va at the location of the left-hand insulating plate (x = +d1) with the potential Vb at the location of the right-hand plate (x = d1 + d2). ANSWER (Va

Explanation / Answer

Since;

V=-int(E.dr)

Va=-(((Sigma)b/2*epsilon)*d2+((Sigma)Lhs/2*epsilon)*(2*d1+w)+((Sigma)Rhs/2*epsilon)*2*d1 )

Vb=-(((Sigma)a/2*epsilon)*d2+((Sigma)Lhs/2*epsilon)*(2*d1+d2+w)+((Sigma)Rhs/2*epsilon)*(2*d1+d2))

Now,

Vb-Va=-((((Sigma)a-(Sigma)b)/2*epsilon)*d2+((Sigma)Lhs/2*epsilon)*(d2)+((Sigma)Rhs/2*epsilon)*(d2))

==>Vb-Va=(5/epsilon)*d2

Apositive quantity hence Vb>Va

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