Relation Between Field and Potential MCQ Quiz - Objective Question with Answer for Relation Between Field and Potential - Download Free PDF

Solution:

The potential field is given as:

$V=\sin x-\cos x$

The relationship between the potential $V$ and the charge density $\rho$ in vacuum is given by Poisson's equation:

${\mathrm{\nabla }}^{2}V=-\frac{\rho }{{ϵ}_0}$

 Calculate the Laplacian of $V$.

The Laplacian in one dimension is:

${\mathrm{\nabla }}^{2}V=\frac{{\partial }^{2}V}{\partial x^2}$

First derivative of $V$:

$\frac{\partial V}{\partial x}=\cos x+\sin x$

Second derivative of $V$:

$\frac{{\partial }^{2}V}{\partial x^2}=-\sin x+\cos x$

 Use Poisson's equation to find $\rho$:

${\mathrm{\nabla }}^{2}V=-\frac{\rho }{{ϵ}_0}$

Substitute the value of ${\mathrm{\nabla }}^{2}V$:

$-\sin x+\cos x=-\frac{\rho }{{ϵ}_0}$

Rearranging:

$\rho ={ϵ}_0(\sin x-\cos x)$

at x=π/4 

$$\begin{gathered} \rho ={ϵ}_0(\sin \pi /4-\cos \pi /4) \\ \rho ={ϵ}_0(1/\sqrt{2}-1/\sqrt{2}) \\ \rho =0 \end{gathered}$$

The correct option is 2): The volume charge density is zero at x=π/4 :

At point A, let the electric field be ‘E’ $\mathrm{k}\mathrm{V}/\mathrm{c}\mathrm{m}$

$\Rightarrow \left(\mathrm{E}-20\right)=\frac{\left(40\right)-\left(20\right)}{\left(5\times {10}^{-4}\right)}\left(\mathrm{x}-0\right)$

The above expression is for the electric field at distance ‘$\mathrm{x}$’ from point A. 

$\Rightarrow \mathrm{E}=4\times {10}^4\mathrm{x}+20$

${\mathrm{V}}_{\mathrm{A}\mathrm{B}}={\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=-\underset{\mathrm{A}}{\overset{\mathrm{B}}{\int }}\overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{l}}$

Notice that $\overrightarrow{\mathrm{E}}$ and $\mathrm{d}\overrightarrow{\mathrm{l}}$ are n the opposite direction. So, the dot product will produce an additional negative sign.

$\Rightarrow {\mathrm{V}}_{\mathrm{A}\mathrm{B}}=-\underset{0}{\overset{5\times {10}^{-4}}{\int }}\left(4\times {10}^4\mathrm{x}+20\right)\mathrm{d}\mathrm{x}=(2\times {10}^4\left({\mathrm{x}}^2\right)+20\mathrm{x}){|}_0^{5\times {10}^{-4}}$

$=-\left(50\times {10}^{-4}+100\times {10}^{-4}\right)$

$=-\left(150\times {10}^{-4}\right)\mathrm{k}\mathrm{V}$

${\mathrm{V}}_{\mathrm{A}\mathrm{B}}={\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=-15\mathrm{V}$

Concept:

The electric field is related to the potential as:

$\overrightarrow{E}=-\nabla V$

$\nabla =\frac{\partial }{\partial x}{\hat{a}}_x+\frac{\partial }{\partial y}{\hat{a}}_y+\frac{\partial }{\partial z}{\hat{a}}_z$

Calculation:

We can verify all options with the Maxwell equation $\overrightarrow{E}=-\nabla V$

Option 1: V = xy3 – yz2

$-\nabla V=-(\frac{\partial }{\partial x}(x{y}^3–y{z}^2){\hat{a}}_x+\frac{\partial }{\partial y}(x{y}^3–y{z}^2){\hat{a}}_y+\frac{\partial }{\partial z}(x{y}^3–y{z}^2){\hat{a}}_z)$

= -y3 ax - (3xy2 - z2) ay - 2yz az

Since this is not equal to the given Electric field, this cannot be a valid expression for the given electrostatic potential.

Option 2: V = 2xy3 – xyz2

$-\nabla V=-(\frac{\partial }{\partial x}(2x{y}^3–xy{z}^2){\hat{a}}_x+\frac{\partial }{\partial y}(2x{y}^3–xy{z}^2){\hat{a}}_y+\frac{\partial }{\partial z}(2x{y}^3–xy{z}^2){\hat{a}}_z)$

= -[(2y3 - yz2) ax + (6xy2 - xz2) ay - 2xyz az]

Since this is not equal to the given Electric field, this cannot be a valid expression for the given electrostatic potential.

Option 3: V = y3 + xyz2

$-\nabla V=-(\frac{\partial }{\partial x}(y^3+xy{z}^2){\hat{a}}_x+\frac{\partial }{\partial y}(y^3+xy{z}^2){\hat{a}}_y+\frac{\partial }{\partial z}(y^3+xy{z}^2){\hat{a}}_z)$

= -[yz2 ax + (3y2 + xz2) ay + 2xyz az]

Since this is not equal to the given Electric field, this cannot be a valid expression for the given electrostatic potential.

Option 4: V = 2xy3 – 3xyz2

$-\nabla V=-(\frac{\partial }{\partial x}(2x{y}^3–3xy{z}^2){\hat{a}}_x+\frac{\partial }{\partial y}(2x{y}^3–3xy{z}^2){\hat{a}}_y+\frac{\partial }{\partial z}(2x{y}^3–3xy{z}^2){\hat{a}}_z)$

= -[(2y3 - 3yz2)ax + (6xy2 - 3xz2) ay - 6xyz az]

= -(2y3 - 3yz2)ax - (6xy2 - 3xz2) ay + 6xyz az

Since E = - ∇ V for this case, this is a valid expression for the given electrostatic potential.

Solution:

The potential field is given as:

$V=\sin x-\cos x$

The relationship between the potential $V$ and the charge density $\rho$ in vacuum is given by Poisson's equation:

${\mathrm{\nabla }}^{2}V=-\frac{\rho }{{ϵ}_0}$

 Calculate the Laplacian of $V$.

The Laplacian in one dimension is:

${\mathrm{\nabla }}^{2}V=\frac{{\partial }^{2}V}{\partial x^2}$

First derivative of $V$:

$\frac{\partial V}{\partial x}=\cos x+\sin x$

Second derivative of $V$:

$\frac{{\partial }^{2}V}{\partial x^2}=-\sin x+\cos x$

 Use Poisson's equation to find $\rho$:

${\mathrm{\nabla }}^{2}V=-\frac{\rho }{{ϵ}_0}$

Substitute the value of ${\mathrm{\nabla }}^{2}V$:

$-\sin x+\cos x=-\frac{\rho }{{ϵ}_0}$

Rearranging:

$\rho ={ϵ}_0(\sin x-\cos x)$

at x=π/4 

$$\begin{gathered} \rho ={ϵ}_0(\sin \pi /4-\cos \pi /4) \\ \rho ={ϵ}_0(1/\sqrt{2}-1/\sqrt{2}) \\ \rho =0 \end{gathered}$$

The correct option is 2): The volume charge density is zero at x=π/4 :

CONCEPT:

• Electric Potential (V): The amount of work done in bringing a unit positive charge from infinity to a given point in an electric field is called electric potential.
• Electric Field (E): The electric force per unit positive charge at point is called electric field. It is a vector quantity.
• Electric Force at a point: The electric force at a point for a charge is the product of charge and electric field.

$$$\overrightarrow{F}=q\overrightarrow{E}$

• The relation between Field and Potential: The relationship between field and potential is given as 

$\overrightarrow{E}=\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}$

CALCULATION:

Given

Potential at point V (x, y, z) = 6x - 8xy - 8y + 6yz

Charge = 2 C

Now,

$\frac{\partial V}{\partial x}=\frac{\partial (6x-8xy-8y+6yz)}{\partial x}$

⇒  $\frac{\partial V}{\partial x}$= 6x - 8y -- (1)

In similar way

 $\frac{\partial V}{\partial y}=-8x-8+6z$   --- (2)

$\frac{\partial V}{\partial z}=6y$ -- (3)

Putting (1), (2), and (3) in the expression for the electric field we get

$\overrightarrow{E}=(6x-8y)\hat{i}+(-8x-8+6z)\hat{j}+6y\hat{k}$  ------ (4)

Now, the given coordinate of the point is (1, 1, 1)

Putting these values in Eq (4) we get 

$\overrightarrow{E}=[6(1)-8(1)]\hat{i}+[-8(1)-8+6(1)]\hat{j}+6(1)\hat{k}$

⇒ $\overrightarrow{E}=-2\hat{i}-10\hat{j}+6\hat{k}$

$\overrightarrow{F}=q\overrightarrow{E}=2\overrightarrow{E}=-4\hat{i}-20\hat{j}+12\hat{k}$

The magnitude of this force

$\left|\overrightarrow{F}\right|=\surd (-4{)}^2+(-20{)}^2+(12{)}^2$

= √ 560

= 4 √ 35 N

Important Points

 ∂ operator is used for partial differentiation. If the partial differentiation is with respect to x, then other variables like y and z are considered constant.

For example, f(x, y, z) = 3xy - 5y + xz

$\frac{\partial (f(x,y,z))}{\partial x}=3y+xz$

That is differentiating with respect to the only x and keeping y and z constant.

Moving along (0, 5, 0) → (2, 5, 0) 

Work done, $W=-Q\int E.dl$ 

$=-Q\int \left(3x^2+y\right)dx+xdy$

$=-Q\left[{\underset{0}{\overset{2}{\int }}\left(3x^2+y\right)dx|}_{y=5}+{\underset{5}{\overset{-1}{\int }}xdy|}_{x=2}\right]$

= -Q [18 - 12] kV

= -(-2 × 10^-6) (6 × 10³)

From third equation of motion, final velocity

$\begin{array}{l}{\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{a}\mathrm{s}\\ \mathrm{A}\mathrm{c}\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n},\mathrm{a}=\frac{\mathrm{q}\mathrm{E}}{\mathrm{m}}\end{array}$ 

And electric field, $\mathrm{E}=\frac{\mathrm{V}}{\mathrm{s}}$ 

Given, $\mathrm{u}=0$, we have

$\begin{array}{l}{\mathrm{v}}^2=2\cdot \left(\frac{\mathrm{q}\mathrm{V}}{\mathrm{m}\mathrm{s}}\right)\mathrm{s}\\ \Rightarrow {\mathrm{v}}^2=\frac{2\mathrm{q}\mathrm{V}}{\mathrm{m}}\\ \Rightarrow \mathrm{v}=\sqrt{\frac{2\mathrm{q}}{\mathrm{m}}}\cdot \sqrt{\mathrm{V}}\end{array}$ 

Comparing with $\mathrm{v}=\mathrm{k}{\mathrm{V}}^{\mathrm{n}}$, we have

$\overrightarrow{\mathrm{E}}=-\overrightarrow{\mathrm{\nabla }}\mathrm{V}=\mathrm{y}\mathrm{z}{\hat{\mathrm{a}}}_{\mathrm{x}}+\mathrm{x}\mathrm{z}{\hat{\mathrm{a}}}_{\mathrm{y}}+\mathrm{x}\mathrm{y}{\hat{\mathrm{a}}}_{\mathrm{z}}$

Total Energy ${\mathrm{w}}_{\mathrm{e}}=\frac{{\epsilon }_0}{2}\int \int \int \overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{E}}\mathrm{d}\mathrm{v}=\frac{{\epsilon }_0}{2}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left({\mathrm{y}}^2{\mathrm{z}}^2+{\mathrm{x}}^2{\mathrm{z}}^2+{\mathrm{x}}^2{\mathrm{y}}^2\right)\mathrm{d}\mathrm{x}\mathrm{d}\mathrm{y}\mathrm{d}\mathrm{z}$

$\Rightarrow {\mathrm{w}}_{\mathrm{e}}=\frac{{\epsilon }_0}{2}\left\{1.\frac{1}{3}.\frac{1}{3}+1.\frac{1}{3}.\frac{1}{3}+1.\frac{1}{3}.\frac{1}{3}\right\}=\frac{{\epsilon }_0}{2}\frac{1}{9}.3=\frac{{\epsilon }_0}{6}$

$\overrightarrow{\mathrm{E}}=-\overrightarrow{\mathrm{\nabla }}\mathrm{V}$

$=-\left[\frac{\partial \mathrm{V}}{\partial \mathrm{r}}{\hat{\mathrm{a}}}_{\mathrm{r}}+\frac{1}{\mathrm{r}}\frac{\partial \mathrm{V}}{\partial \theta }{\hat{\mathrm{a}}}_{\theta }+\frac{1}{\mathrm{r}\mathrm{s}\mathrm{i}\mathrm{n}\theta }\frac{\partial \mathrm{V}}{\partial \varphi }{\hat{\mathrm{a}}}_{\varphi }\right]$

$=-\left[\frac{-10}{{\mathrm{r}}^2}\mathrm{s}\mathrm{i}\mathrm{n}\theta \mathrm{c}\mathrm{o}\mathrm{s}\varphi {\hat{\mathrm{a}}}_{\mathrm{r}}+\frac{10}{{\mathrm{r}}^2}\mathrm{c}\mathrm{o}\mathrm{s}\theta \mathrm{c}\mathrm{o}\mathrm{s}\varphi {\hat{\mathrm{a}}}_{\theta }+\frac{1}{\mathrm{r}\mathrm{s}\mathrm{i}\mathrm{n}\theta }\frac{10}{\mathrm{r}}\mathrm{s}\mathrm{i}\mathrm{n}\theta \left(-\mathrm{s}\mathrm{i}\mathrm{n}\varphi \right){\hat{\mathrm{a}}}_{\varphi }\right]$

$\Rightarrow \overrightarrow{\mathrm{E}}=\frac{10}{{\mathrm{r}}^2}\mathrm{s}\mathrm{i}\mathrm{n}\theta \mathrm{c}\mathrm{o}\mathrm{s}\varphi {\hat{\mathrm{a}}}_{\mathrm{r}}-\frac{10}{{\mathrm{r}}^2}\mathrm{c}\mathrm{o}\mathrm{s}\theta \mathrm{c}\mathrm{o}\mathrm{s}\varphi {\hat{\mathrm{a}}}_{\theta }+\frac{10}{{\mathrm{r}}^2}\mathrm{s}\mathrm{i}\mathrm{n}\varphi {\hat{\mathrm{a}}}_{\varphi }$

$\overrightarrow{\mathrm{E}}=\frac{10}{{\mathrm{r}}^2}\left[\mathrm{s}\mathrm{i}\mathrm{n}\theta \mathrm{c}\mathrm{o}\mathrm{s}\varphi {\hat{\mathrm{a}}}_{\mathrm{r}}-\mathrm{c}\mathrm{o}\mathrm{s}\theta \mathrm{c}\mathrm{o}\mathrm{s}\varphi {\hat{\mathrm{a}}}_{\theta }+\mathrm{s}\mathrm{i}\mathrm{n}\varphi {\hat{\mathrm{a}}}_{\varphi }\right]$