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Relation Between Electric Field Intensity And Electric potential

Relation Between Electric Field Intensity And Electric potential :- From the definition, potential difference between two points A and B is given by :

$\displaystyle {{V}_{B}}-{{V}_{A}}=\frac{{{W}_{AB}}}{{{q}_{o}}}$ …..(1)

If the points A and B are very close ( $\displaystyle \overrightarrow{dr}$ ) to each others, then

$\displaystyle {{V}_{B}}-{{V}_{A}}=dV$

$\displaystyle \Rightarrow dV=\frac{{{W}_{AB}}}{{{q}_{o}}}=\frac{\left( \overrightarrow{{{F}_{ext}}} \right).\left( \overrightarrow{dr} \right)}{{{q}_{o}}}$

$\displaystyle \Rightarrow dV=\frac{\left( -{{q}_{o}}\overrightarrow{E} \right).\left( \overrightarrow{dr} \right)}{{{q}_{o}}}$

$\displaystyle dV=-\overrightarrow{E}.\overrightarrow{dr}$ …..(2)

Integrating equation (2) we get,

$\displaystyle \int{dV=-\int{\overrightarrow{E}.\overrightarrow{dr}}}$

$\displaystyle V=-\int\limits_{\infty }^{r}{\overrightarrow{E}.\overrightarrow{dr}}$ …..(3)

In equation (2), the the small displacement ( $\displaystyle \overrightarrow{dr}$ ) and electric field ( $\displaystyle \overrightarrow{E}$ ) can be written as,

$\displaystyle \overrightarrow{dr}=dx\widehat{i}+dy\widehat{j}+dz\widehat{k}$ …..(4)

$\displaystyle \overrightarrow{E}={{E}_{x}}\widehat{i}+{{E}_{y}}\widehat{j}+{{E}_{z}}\widehat{k}$ …..(5)

Using equations (4) and (5) in equation (2) :-

$\displaystyle dV=-\overrightarrow{E}.\overrightarrow{dr}=-\left( {{E}_{x}}\widehat{i}+{{E}_{y}}\widehat{j}+{{E}_{z}}\widehat{k} \right).\left( dx\widehat{i}+dy\widehat{j}+dz\widehat{k} \right)$

$\displaystyle dV=-\left( {{E}_{x}}dx+{{E}_{y}}dy+{{E}_{z}}dz \right)$ …..(6)

Now the total differential of is :-

$\displaystyle dV=\frac{\partial V}{\partial x}dx+\frac{\partial V}{\partial y}dy+\frac{\partial V}{\partial z}dz$ …..(7)

Comparing equations (6) and (7),

$\displaystyle {{E}_{x}}=-\frac{\partial V}{\partial x}\text{, }{{E}_{y}}=-\frac{\partial V}{\partial y},\text{ }{{E}_{z}}=-\frac{\partial V}{\partial z}$

Putting the values of E_x , E_y and E_z in equation (5),

$\displaystyle \overrightarrow{E}=-\frac{\partial V}{\partial x}\widehat{i}-\frac{\partial V}{\partial y}\widehat{j}-\frac{\partial V}{\partial z}\widehat{k}$

$\displaystyle \overrightarrow{E}=-\left( \frac{\partial }{\partial x}\widehat{i}+\frac{\partial }{\partial y}\widehat{j}+\frac{\partial }{\partial z}\widehat{k} \right)V$

$\displaystyle \Rightarrow \overrightarrow{E}=-\overrightarrow{\nabla }\text{ V}$ …..(8)

Equation (8) shows that the electric field is the negative gradient of the electric potential.

Here $\displaystyle \overrightarrow{\nabla }=del\text{ }operator$ .

$\displaystyle \vec{\nabla }\text{ }\!\!~\!\!\text{ }=\left( \frac{\partial }{\partial x}\hat{i}+\frac{\partial }{\partial y}\hat{j}+\frac{\partial }{\partial z}\hat{k} \right)$

It is denoted by the nabla symbol is a vector differential operator used in vector calculus.

Special Cases :

From equation (2),

$\displaystyle dV=-\vec{E}.\overrightarrow{dr}=-Edrcos\theta$

(i). If we move in the direction of electric field (θ = 0°)

$\displaystyle dV=-Edrcos0=-Edr=-ve$

In the direction of electric field ( $\displaystyle {\vec{E}}$ ), electric potential decreases.

Further, the magnitude of electric field is given by change in magnitude of potential per unit displacement normal to the equipotential surface at the point. this is called potential gradient, i.e.,

$\displaystyle E=-\frac{dV}{dr}$

(ii). If we move in opposite direction of electric field (θ = 180°)

$\displaystyle dV=-Edrcos180=-Edr\left( -1 \right)=Edr=+ve$

In opposite direction of electric field ( $\displaystyle {\vec{E}}$ ), electric potential increases.

(iii). If we move perpendicular to electric field (θ = 90°)

$\displaystyle dV=-Edrcos90=-Edr\left( 0 \right)=0$

$\displaystyle \Rightarrow V=constant$

If we move perpendicular to electric field, electric potential remains same.

Example 1.

Electric potential at a point in space is given by V = 3x + 4y + 5z . Find an expression for electric field intensity. Is the electric field uniform of non-uniform ? Find the force experienced by a charge of 2C placed at (1,1,1).

Solution :-

$\displaystyle \vec{E}=-\left( \frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\widehat{k} \right)=-\left[ \frac{\partial }{\partial x}(3x+4y+5z)\hat{i}+\frac{\partial }{\partial y}(3x+4y+5z)\hat{j}+\frac{\partial }{\partial z}(3x+4y+5z)\widehat{k} \right]$

$\displaystyle \Rightarrow \vec{E}=-\left[ \frac{\partial }{\partial x}(3+0+0)\hat{i}+\frac{\partial }{\partial y}(0+4+0)\hat{j}+\frac{\partial }{\partial z}(0+0+5)\widehat{k} \right]$

$\displaystyle \therefore \vec{E}=-3\hat{i}-4\hat{j}-5\widehat{k}$

This is the required expression for electric field intensity. Here we find that this expression is independent of position (x,y,z) , hence it is a uniform electric field.

Force on 2C charge :

$\displaystyle F=q\left| {\vec{E}} \right|=2\sqrt{9+16+25}=10\sqrt{2}Newton$

Example 2.

Electric potential at a point in space is given by V = x² + y² + z² . Find an expression for electric field intensity. Is the electric field uniform of non-uniform ?

Solution :-

$\displaystyle \vec{E}=-\left( \frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\widehat{k} \right)=-\left[ \frac{\partial }{\partial x}({{x}^{2}}+{{y}^{2}}+{{z}^{2}})\hat{i}+\frac{\partial }{\partial y}({{x}^{2}}+{{y}^{2}}+{{z}^{2}})\hat{j}+\frac{\partial }{\partial z}({{x}^{2}}+{{y}^{2}}+{{z}^{2}})\widehat{k} \right]$

$\displaystyle \Rightarrow \vec{E}=-\left[ \frac{\partial }{\partial x}(2x+0+0)\hat{i}+\frac{\partial }{\partial y}(0+2y+0)\hat{j}+\frac{\partial }{\partial z}(0+0+2z)\widehat{k} \right]$

$\displaystyle \therefore \vec{E}=-2x\hat{i}-2y\hat{j}-2z\hat{k}$

This is a non – uniform electric field as it depends on position (x,y,z) .

Example 3.

Electric potential at a point in space is given by V = xy + yz + zx . Find an expression for electric field intensity. Is the electric field uniform of non-uniform ?

Solution :-

$\displaystyle \vec{E}=-\left( \frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{z} \right)=-\left[ \frac{\partial }{\partial x}(xy+yz+zx)\hat{i}+\frac{\partial }{\partial y}(xy+yz+zx)\hat{j}+\frac{\partial }{\partial z}(xy+yz+zx)\hat{z} \right]$

$\displaystyle \Rightarrow \vec{E}=-\left[ \frac{\partial }{\partial x}(y+0+z)\hat{i}+\frac{\partial }{\partial y}(x+z+0)\hat{j}+\frac{\partial }{\partial z}(0+y+x)\hat{z} \right]$

$\displaystyle \therefore \vec{E}=-(y+z)\hat{i}-(x+z)\hat{j}-(y+x)\widehat{k}$

This is a non-uniform electric field.

Example 4.

In the followings figures, how electric potential changes as we move from point A to point B :-

Solution :-

(i) As we move from A to B first electric potential decreases up to the mid point O and then it increases from point O to B.

(ii) From A to B first electric potential increases and then decreases.

(iii) From A to B electric potential decreases continuously.

Example 5.

Compare the electric potential of points A, B, C and D in the following figures :

Solution :-

Figure (a)

As in the direction of electric field, electric potential decreases, so

V_A > V_B > V_C

Figure (b)

V_A > V_B= V_D > V_C

Example 6.

A uniform electric field is along x–axis . The potential difference V_A– V_B = 10 V is between two points A (2m , 3m) and B (4m, 8m). Find the electric field intensity.

Solution :-

As $\displaystyle E=-\frac{dV}{dr}$ , here -ve sign indicates that electric potential decreases in the direction of electric field. In magnitude we can write :

$\displaystyle E=\frac{\Delta V}{\Delta r}$

Here ΔV = 10 V and Δr = 4 – 2 = 2m (because electric field is along x-axis only), so

$\displaystyle E=\frac{10}{2}=5\text{ }V/m$

Example 7.

For given electric field $\displaystyle \overrightarrow{E}=2x\widehat{i}+3y\widehat{j}$ , find the potential at (x, y) if V at origin is 5 volts.

Solution :-

As $\displaystyle dV=-\overrightarrow{E}.\overrightarrow{dr}$ , so

$\displaystyle \int\limits_{5}^{V}{dV}=-\int\limits_{(0,0)}^{(x,y)}{\vec{E}.\overrightarrow{dr}}=-\int\limits_{(0,0)}^{(x,y)}{\left( {{E}_{x}}\widehat{i}+{{E}_{y}}\widehat{j} \right).\left( dx\widehat{i}+dy\widehat{j} \right)}$