Relation Between Electric Field And Electric Potential

The concepts of electric field and electric potential are crucial for understanding the fundamentals of electrostatics, which focuses on stationary electric charges and their interactions. While these two ideas are interrelated, they serve different purposes in our understanding of electric phenomena. The electric field allows us to visualize how electric charges affect their environment, making it easier to predict how other charges will react when subjected to electric forces. In contrast, electric potential is essential for grasping how energy changes take place in electrostatic systems. Together, these concepts provide valuable insights into the behavior of electric charges and their effects in various contexts.

1.0Definition Of Electric Field

• The space around a charge or charge distribution, in which another charge experiences an electric force, is called an electric field.
• Every charge has its own electric field.
• Electric field at a point can be characterized by

1. A vector function of position, called intensity of electric field ( )

2. A scalar function of position, called electrostatic potential(V).

3. Graphically, with help of electric field lines (EFL)

•    It is defined as the net force experienced by a unit positive test charge.

$\vec{E}=\frac{\vec{F}}{q_0}$

SI Unit : $\frac{N}{C}\text{ or }\frac{V}{m}$

Dimensional  Formula: $\left[M^1{L}^1{T}^{-3}{A}^{-1}\right]$

2.0Representation of Electric Field

3.0Electric Potential

• In an electrostatic field, the electric potential (due to some source charges) at a point P is defined as the work done by an external agent in taking a unit point positive charge from a point of reference (generally taken at infinity) to that point P without changing its kinetic energy.
• If ${\left(W_{\infty \to P}\right)}_{ext}$ is the work required in moving a point charge q from infinity to a point P ,the electric potential at point P is given by

$V_P=\frac{{\left(w_{\infty \to p}\right)}_{ext}}{q}(\Delta K=0)$

• It is scalar quantity
• SI Unit of potential is $\text{ Volt }=\frac{\text{ Joule }}{\text{ Coulomb }}$
• Dimensional Formula $\left[M^1{L}^2{T}^{-3}{A}^{-1}\right]$
• Potential can be positive,negative and even zero.Hence charge must be taken with sign.
• Electric potential always decreases in the direction of the electric field.

$V_P={\left(\frac{W_{\text{agent }}}{q_0}\right)}_{\Delta K=0}$

$V_P={\left(-\frac{W_{\text{electric }}}{q_0}\right)}_{\Delta K=0}$

•  Electric potential at a point is also equal to the negative of the work done by the electric field in taking the point charge from reference point (i.e. infinity) to that point.

$V_P=-\frac{{\left(W_{\infty \to P}\right)}_{ext}}{q}(\Delta K=0)$

4.0Electric Potential Due To A Point Charge

The work accomplished in moving a unit positive charge from infinity to a specific point against the electric field, without altering its kinetic energy, is defined as the electric potential at that point. 

$W_{\infty \to P}=\frac{KQ{q}_0}{r}$

$V_P=\frac{W_{\infty \to P}}{q_0}=\frac{KQ{q}_0}{q_{0}r}=\frac{kQ}{r}$

• Potential due to Positive Charge $V_P=+\frac{kQ}{r}$
• Potential due to Negative Charge $V_P=-\frac{kQ}{r}$
• With reference potential is zero at infinity.

5.0Relation Between Electric Field And Electric Potential

$F=-\frac{dU}{dr}$

Dividing both sides by q

$\frac{F}{q}=-\frac{d}{dr}\left(\frac{U}{q}\right)\Rightarrow E=-\frac{dV}{dr}(F=qE\text{ And }U=qV)$

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

• So, the negative gradient of electric potential is equal to electric field.
• Since the electric field is a negative gradient of potential, thus potential decreases in the direction of.

Finding E when $\text{ V }$ is given:

• Gradient $\to$ Rate of change with respect to distance
• In cartesian system,as

$\vec{E}=-\vec{\nabla }Vand\nabla =\frac{\partial }{\partial x}\hat{i}+\frac{\partial }{\partial y}\hat{j}+\frac{\partial }{\partial x}\hat{k}$

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

$\vec{E}=E_x\hat{i}+E_y\hat{j}+E_z\hat{k}$

$E_x=-\frac{\partial V}{\partial x}$

$E_y=-\frac{\partial V}{\partial y}$

$E_z=-\frac{\partial V}{\partial z}$

Finding V When E is given:

As we know

$E=-\frac{dV}{dr}\Rightarrow dV=-Edr\Rightarrow dV=-\vec{E}\cdot \overrightarrow{dr}$

$\overrightarrow{dr}=dx\hat{i}+dy\hat{j}+dz\hat{k}$

On integrating,

$\Delta V=V_P-V_{ref}=-{\int }_{ref}^P\vec{E}\cdot \overrightarrow{dr}$

$V_P-V_{ref}=-{\int }_{x_{ref}}^{x_p}{E}_{x}dx-{\int }_{y_{ref}}^y{E}_{y}dy-{\int }_{Z_{ref}}^{Z_p}{E}_{z}dz$

6.0Difference Between Electric Field And Electric Potential

7.0Sample Questions On Relation Between Electric Field And Electric Potential.

Q-1. If $V=8x^2$, find an electric field in the region?

Solution:

$E_x=-\frac{\partial V}{\partial x}=-16x$

$\vec{E}=E_x\hat{i}+E_y\hat{j}+E_z\hat{k}\left(E_y\text{ and }{E}_z\text{ is zero }\right)$

$\vec{E}=-16x(-\hat{i})$

Q-2. If $V=x^{2}y+y^{2}xz+5$ then find $\vec{E}$ at (1,1,0) 

Solution:

$E_x=-\frac{\partial V}{\partial x}=-\left[2xy+y^{2}z\right]$

$E_y=-\frac{\partial V}{\partial y}=-\left[x^2+2yxz\right]$

$E_z=-\frac{\partial V}{\partial z}=-\left[0+y^{2}x\right]$

At (1,1,0) $\Rightarrow E_x=-2\hat{i},E_y=-\hat{j},E_z=-\hat{k}$

$\Rightarrow \vec{E}=-2\hat{i}-\hat{j}-\hat{k}$

Q-3. If $\vec{E}=\frac{40}{x^2}(\hat{i})N/m$, then find the potential difference between the points x=2 m and x=4 m.

Solution:

$\int dV=-\int Edx\mathrm{Cos}\theta$

${\int }_{V_A}^{V_B}dV=-{\int }_{x=2}^{x=4}Edx\mathrm{Cos}{0}^{\circ }$

$\left(V_B-V_A\right)=-40{\left[-\frac{1}{x}\right]}_2^4$

$\left(V_B-V_A\right)=+40\left[\frac{1}{4}-\frac{1}{2}\right]=-\frac{40}{4}=-10\mathrm{V}$

$V_A-V_B=+10\mathrm{V}$