Two sheets of charge lie parallel to one another and the positive sheet is \"abo

Question

Two sheets of charge lie parallel to one another and the positive sheet is "above" the negative sheet as depicted in the sketch below. The surface charge density of the positive sheet is 17.4 x 10-6 micro-coulomb/m2, while the surface charge density of the negative sheet is -4.3 x 10-6 micro-coulomb/m2. (Note that due to the sheets not having the same magnitude of charge density, this configuration is NOT a parallel plate capacitor.) How much does the electric potential change due to both sheets from 0.21 meters above the positive sheet to 4.31 meters above the positive sheet in MV (1 x 106 V)? If negative include a negative sign.

Explanation / Answer

electric field = ( 17.4*10^-9 + 4.3*10^-9)/ 2*8.85*10^-12 ( direction is towards negative plate)

E = 1225.9887 N/C

voltage difference is positive opposite to the field direction

hence voltage diffference =1225.9887*(4.31 - .21)

=5026.55367

= 0.053 MV

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