Question

The electric potential existing in space is V(x, y, z)=A(xy+yz+zx). (a) Write the dimensional formula of A. (b) Find the expression for the electric field. (c) If A is 10 SI units, find the magnitude of the electric field at (1 m, 1 m, 1 m).

Answer 1

Explanation:

Given

There is Electric potential in space   $\begin{equation}V(\mathrm{x}, \mathrm{y}, \mathrm{z})=\mathrm{A}(\mathrm{xy}+\mathrm{yz}+\mathrm{zx})\end$

(a) Write A's dimension formula.

    $\begin{equation}\begin{aligned}A &=\frac{v o l t}{m^{2}} \\A &=\frac{M L^{2} T^{-2}}{I T L^{2}} \\A &=M T^{-3} I^{-1}\end{aligned}\end$

(b) Find the electric field expression

Let E be the electric field.

$\begin{equation}E=-\frac{\delta V_{i}}{\delta x}-\frac{-\delta V_{j}}{\delta y}-\frac{-\delta V_{k}}{\delta z}\end$

$\begin{equation}E=-\left[\frac{\delta}{\delta x} \mathrm{A}(\mathrm{xy}+\mathrm{yz}+\mathrm{zx})+\frac{\delta}{\delta y} \mathrm{A}(\mathrm{xy}+\mathrm{yz}+\mathrm{zx})+\frac{\delta}{\delta z} \mathrm{A}(\mathrm{xy}+\mathrm{yz}+\mathrm{zx})\right.$

$\begin{equation}E=-[(A y+A z) i+(A x+A z) j+(A y+A x) k\end$

$\begin{equation}E=-A(y+z) i-\mathrm{A}(A+z) j-A(y+x) k\end$

(c) Consider the magnitude of the electric field (1 m, 1 m, 1 m) if A is 10 SI units.

A = 10 si unit

r = (1 m, 1 m, 1 m)

$\begin{equation}E=-10(1+1) i-10(1+1) j-10(1+1) k\end$

$\begin{equation}E=-10(2) i-10(2) j-10(2) k\end$

$\begin{equation}E=-20 i-20 j-20 k\end$

Magnitude of electric field,

$\begin{equation}E=(-20 i)+(-20 j)+(-20 k)\end$

$\begin{equation}E=\sqrt{(-20 i)^{2}+(-20 j)^{2}+(-20 k)^{2}}$

$\begin{equation}E=\sqrt{\left(-400 i^{2}\right)+\left(-400 j^{2}\right)+\left(-400 k^{2}\right)}$

Where  

$i^2$ = -1

$j^2$ = -1

$k^2$ = -1

$\begin{equation}E=\sqrt{(-400(-1))+(-400(-1))+(-400(-1))}\end$

$\begin{equation}E=\sqrt{400+400+400}\e$

$\begin{equation}E=\sqrt{1200}\end$

$\begin{equation}E=34.64 \approx 35 \frac{N}{C}\end$

Answer 2

Answer:

• Dimensional Formula:- M T⁻³ I⁻¹

• Electric Field:- $\sf - A \bigg[(y+z)\hat{i}+(x+z)\hat{j}+(x+y)\hat{k}\bigg]$

• Magnitude :- 20 √(3) V/m

Given:

1. V (x,y,z) = A (xy + yz + zx)
2. A = 10
3. Point given (1 m, 1 m, 1 m)

Explanation:

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Dimensional Formula:

As x, y and z are lengths. Therefore, from the principle of homogeneity the (xy + yz + zx) will be x × y = m².

Hence,

$\longrightarrow\sf V=A\times (Length)^{2}$

Substituting the voltage dimensions,

$\longrightarrow\sf M\;L^{2}\;T^{-3}\;I^{-1}=\Big[A\Big]\times L^{2}\\\\\\\longrightarrow\sf \Big[A\Big] =\dfrac{M\;L^{2}\;T^{-3}\;I^{-1}}{L^{2}}\\\\\\\longrightarrow\sf \Big[A\Big] = M\;T^{-3}\;I^{-1}\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf \Big[A\Big] = M\;T^{-3}\;I^{-1}}}}}$

∴ Dimensional Formula will be M T⁻³ I⁻¹.

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Given relation,

⇒ V(x, y, z) = A (xy + yz +zx)

From the formula we know,

$\bigstar\;\underline{\boxed{\sf E=-\Bigg[\dfrac{\partial V}{\partial x}\;\hat{i}+\dfrac{\partial V}{\partial y}\;\hat{j}+\dfrac{\partial V}{\partial z}\;\hat{k}\Bigg]}}$

Substituting the values,

$\longrightarrow\sf E=-A\Bigg[\dfrac{\partial (xy+yz+zx)}{\partial x}\;\hat{i}+\dfrac{\partial (xy+yz+zx)}{\partial y}\;\hat{j}+\dfrac{\partial (xy+yz+zx)}{\partial z}\;\hat{k}\Bigg]\\\\\\\longrightarrow\sf E=-A\Bigg[(y-0+z)\;\hat{i}+(x+z+0)\;\hat{j}+(0+y+x)\;\hat{k}\Bigg]\\\\\\\longrightarrow\sf E=-A\Bigg[(y+z)\;\hat{i}+(x+z)\;\hat{j}+(x+y)\;\hat{k}\Bigg]\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf E=-A\bigg[(y+z)\;\hat{i}+(x+z)\;\hat{j}+(x+y)\;\hat{k}\bigg]\;Vm^{-1}}}}}$

∴ We got the Electric Field.

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Now, Finding the magnitude at (1 m, 1 m, 1 m)

Substituting the values in the above equation,

$\longrightarrow\sf E=-A\Bigg[(y+z)\;\hat{i}+(x+z)\;\hat{j}+(x+y)\;\hat{k}\Bigg]\\\\\\\longrightarrow\sf E=-A\Bigg[(1+1)\;\hat{i}+(1+1)\;\hat{j}+(1+1)\;\hat{k}\Bigg]\\\\\\\longrightarrow\sf E=-A\bigg[2\;\hat{i}+2\;\hat{j}+2\;\hat{k}\bigg]\\\\\\\longrightarrow\sf E=-10\times (2\;\hat{i}+2\;\hat{j}+2\;\hat{k})\\\\\\\longrightarrow\sf E=-20\;\hat{i}-20\;\hat{j}-20\;\hat{k}\\\\\\\longrightarrow\boxed{\sf E=-20\;\hat{i}-20\;\hat{j}-20\;\hat{k}}$

Now, Lets find the Magnitude.

$\displaystyle\longrightarrow\sf |E|=\sqrt{(E_x)^{2}+(E_y)^{2}+(E_z)^{2}}$

Substituting the values,

$\displaystyle\longrightarrow\sf |E|=\sqrt{(-20)^{2}+(-20)^{2}+(-20)^{2}}\\\\\\\longrightarrow\sf |E|=\sqrt{400+400+400}\\\\\\\longrightarrow\sf |E|=\sqrt{400\times 3}\\\\\\\longrightarrow\sf |E|=20\sqrt{3}\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf |E|=20\sqrt{3}\;Vm^{-1}}}}}$

∴ The Magnitude of electric field 20 √(3) V/m.

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