Astraight nonconducting plastic wire 9.5 cm long carries a charge density of 100 nc/m distributed uniformly along its length. it is lying on a horizontal tabletop. a. find the magnitude and direction of the electric field this wire produces at a point 4.50 cm directly above its midpoint. b. if the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 4.50 cm directly above its center.

Explanation:

Given that,

Length of wire = 9.5 cm =

Charge density = 100 NC/m

Distance = 4.50 cm

(A). We need to calculate the  the magnitude and direction of the electric field

Using formula of magnetic field

Here, x = distance from mid point of the wire

Put the value into the formula

The electric field is .

The direction of electric field is upward.

(B). If the wire is now bent into a circle lying flat on the table,

We need to calculate the the magnitude and direction of the electric field

Using formula of electric field

Put the value into the formula

The electric field is

The direction of electric field is upward.

Hence, This is the required solution.

  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