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Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by a distance of 50.0 mm , and the potential difference between them is 345 V .
a. What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?
b. What is the magnitude of the force this field exerts on a particle with a charge of 2.00 nC ?
c. Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.
d. Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

Respuesta :

We are given big parallel metallic plates aa and bb bring contrary expenses of same significance are separated through a distance dd = 45.zero mm = zero.1/2 m wherein the capacity distinction among the 2 plates is V_V ab = 360 V

  • Required(a) The electric powered area among the 2 plates EE(b) The pressure FF exerted because of the electrical area.(c) The paintings performed WW(d) Compare the paintings performed fee to the extrade in capacity energy.
  • Our trouble is associated with instance 23.9, wherein the electrical area among parallel plates is associated with the capacity distinction among the 2 plates through the following equation
  • E = dfrac{V_}E= dV ab
  • Where for a given capacity distinction V_V ab , the smaller the space dd among the 2 plates, the more the significance EE of the elec­tric area.
  • Now we are able to plug our values for V_V ab and dd into equation (1) to get EE begin E &= dfrac{V_} &= dfrac}{zero.1/2 ,text} &= eight instances 10^ ,text &= boxed{eight instances 10^ ,text} endE = dV ab = zero.045m360V =eight×10 3 V/m= eight×10 3 N/C
  • Result2 of 2(a) E = eight.zero instances 10^ ,textE=eight.zero×10 3 N/C

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