Physics, 20.09.2020 19:01 kharmaculpepper

What is VB, the electric potential at point B, located at distance d from one end of the rod (on the x axis)?

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Answer from: jasminecoronetti44

The question is missing some parts. Here is the complete question.

A finite rod of length L has a total charge q, distributed uniformly along its length. The rod lies on the x-axis and is centered at the origin. Thus one endpoint is located at (L/2) and the other is located at (L/2). Define the electrical potential to be zero at an infinite distance away from the rod. Throughout this problem, you may use the constant k in place of the expression 1 / $4\pi\epsilon_{0}$

Part A (image 1): What is VA, the electric potential at point A (see figure below), is located a distance d above the midpoint of the rod on the y-axis? Express your answer in terms of L, d, q and k.

Part B (image 2): What is VB, the electric potential at point B, located at distance d form one end of the rod (on the x-axis)? give your answer in terms of L, d, q and k.

Part A: $V = \frac{kq}{L}[ln(\frac{L/2+\sqrt{(L/2)^{2}+y^{2}} }{-L/2+\sqrt{(-L/2)^{2}+y^{2}} }) ]$

Part B: $V=\frac{kq}{L} ln(\frac{L}{d} )$

Explanation: Electric Potential (V) is the amount of work done per unit charge to move a charge from point A to B.

For a finite rod with charge uniformly distributed along its length

$V=\frac{1}{4\pi\epsilon_{0}} \int\ {\frac{\lambda}{r} } \,dl$

where

λ is charge density and, in this case, is constant: $\lambda=\frac{q}{L}$

dl is differential of the rod

r is the distance the point is from the rod

Part A: $r = \sqrt{(a^{2}+y^{2})}$

$V=k \int\ {\frac{\lambda}{\sqrt{(a^{2}+y^{2})}} } \,da$

$V=k\lambda \int\ {\frac{1}{{\sqrt{(a^{2}+y^{2})}}} } } \,da$

$V=\frac{kq}{L}\int\limits^\frac{L}{2} _\frac{-L}{2} {\frac{1}{\sqrt{(a^{2}+y^{2})} } } \, da$

$V=\frac{kq}{L}ln(\frac{L/2+\sqrt{(L/2)^{2}+y^{2}} }{-L/2+\sqrt{(-L/2)^{2}+y^{2}} } )$

At a point located at y-axis, electric potential is $V=\frac{kq}{L}ln(\frac{L/2+\sqrt{(L/2)^{2}+y^{2}} }{-L/2+\sqrt{(-L/2)^{2}+y^{2}} } )$

Part B: r = x

$V=k \int\limits^L_d {\frac{\lambda}{x} } \, dx$

$V=k \frac{q}{L} \int\limits^L_d {\frac{1}{x} } \, dx$

$V=k \frac{q}{L} ln(L - d)$

$V= \frac{kq}{L} ln(\frac{L}{d} )$

At a point d from one end, electric potential is $V= \frac{kq}{L} ln(\frac{L}{d} )$

What is VB, the electric potential at point B, located at distance d from one end of the rod (on the

Answer from: Quest

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

Answer from: Quest

Ithink the answers is b i hope it my friend message me if i’m wrong and i’ll change my answer and fix it for you

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