Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?

475 N/C

Explanation:

As we know that, the electric field in parallel plate capacitor is same (constant) throughout. And is potential gradient.

So, Electric field is given by

Electric field = potential gradient

Here, the potential change is 3.8V and the distance from negative plate to positive plate is 1.6 cm. The potential from negative plate to the center is (1.6/2)cm i.e., 0.8 cm.

But we have to take distance in SI units So, distance=

So, Electric field is

So, electric field is 475 Volts per meter.

Note : Also we can say 475 Newtons per coulomb

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

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