A -12 nC charrge is located at the origin. a) What

b2sonicxh 2022-01-12 Answered
A -12 nC charrge is located at the origin.
a) What is the strength of the electric field at the position \(\displaystyle{\left({x},{y}\right)}={\left({0}{c}{m},{5.0}{c}{m}\right)}?\)
b)What is the strength of the electric field at the position \(\displaystyle{\left({x},{y}\right)}={\left(-{5.0}{c}{m},-{5.0}{c}{m}\right)}?\)
c)What is the strength of the electric field at the position \(\displaystyle{\left({x},{y}\right)}={\left(-{5.0}{c}{m},{5.0}{c}{m}\right)}?\)

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Expert Answer

kaluitagf
Answered 2022-01-13 Author has 5757 answers

Step 1
The charge located at the origin is given as,
\(\displaystyle{q}=-{12}{n}{C}\)
\(\displaystyle=-{12}{n}{C}\times{\frac{{{10}^{{-{9}}}{C}}}{{{1}{n}{C}}}}\)
\(\displaystyle=-{12}\times{10}^{{-{9}}}{C}\)
Step 2
(a)
The position at which electric field required is given as,
\(\displaystyle{\left({x},{y}\right)}={\left({0}{c}{m},{5}{c}{m}\right)}\)
The distance of the required position from the origin can be determined as,
\(\displaystyle{r}=\sqrt{{{x}^{{{2}}}+{y}^{{{2}}}}}\)
\(\displaystyle=\sqrt{{{\left({0}{c}{m}\right)}^{{{2}}}+{\left({5}{c}{m}\right)}^{{{2}}}}}\)
\(\displaystyle={5}{c}{m}\times{\frac{{{10}^{{-{2}}}{m}}}{{{1}{c}{m}}}}\)
\(\displaystyle={0.05}{c}{m}\)
The electric field strength at the required location can be determined as,
\(E=\frac{k\mid q\mid}{r^{2}}\)
\(=\frac{(9\times10^{9}N.m^{2}/C^{2})\mid-12\times10^{-9}C\mid}{(0.05m)^{2}}\)
\(\displaystyle={43200}\frac{{N}}{{C}}\)

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MoxboasteBots5h
Answered 2022-01-14 Author has 4180 answers

Step 3
(b)
The position at which electric field required is given as,
\(\displaystyle{\left({x},{y}\right)}={\left(-{5}{c}{m},-{5}{c}{m}\right)}\)
The distance of the required position from the origin cam be determined as,
\(\displaystyle{r}=\sqrt{{{x}^{{{2}}}+{y}^{{{2}}}}}\)
\(\displaystyle=\sqrt{{{\left(-{5}{c}{m}\right)}^{{{2}}}+{\left(-{5}{c}{m}\right)}^{{{2}}}}}\)
\(\displaystyle={7.1}{c}{m}\times{\frac{{{10}^{{-{2}}}{m}}}{{{1}{c}{m}}}}\)
\(\displaystyle={0.071}{c}{m}\)
The electric field strength at the required location can be determined as,
\(E=\frac{k\mid q\mid}{r^{2}}\)
\(=\frac{(9\times10^{9}N.m^{2}/C^{2})\mid-12\times10^{-9}C\mid}{(0.071m)^{2}}\)
\(\displaystyle={21424.3}\frac{{N}}{{C}}\)

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star233
Answered 2022-01-15 Author has 0 answers

Step 4
(b)
The position at which electric field required is given as,
\((x,y)=(-5cm,5cm)\)
The distance of the required position from the origin cam be determined as,
\(r=\sqrt{x^{2}+y^{2}} \\=\sqrt{(-5cm)^{2}+(5cm)^{2}} \\=7.1cm\times\frac{10^{-2}m}{1cm} \\=0.071cm\)
The electric field strength at the required location can be determined as,
\(E=\frac{k\mid q\mid}{r^{2}} \\=\frac{(9\times10^{9}N.m^{2}/C^{2})\mid-12\times10^{-9}C\mid}{(0.071m)^{2}} \\=21424.3N/C\)

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