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# Electrical Potential Relation Assignment Help

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Electrostatics - Electrical Potential Relation

**Electrical Potential Relation**

To obtain relation between electric intensity E and electric potential V, let us consider two equipotential surfaces A and B spaced closely as shown in fig. 1 (c) let the potential of A be **VA = V **and potential of B be**VB = (V + dV)** where dV is increase in potential in the direction of electric intensity E normal to A and B.

Suppose dr is perpendicular distance between the two equipotential surfaces. When a unit positive charge is taken along this perpendicular distance from the surface B to the surface A against the electric field.

Work done **WBA = | E | dr**

By definition

**WBA = VA – VB = V – (V + dV) = - dV **

∴ |E| dr = - dV

|E| = - dV / dr

Since** |E| **is positive dV is negative.

Therefore the direction of electric field E is in the direction of decreasing potential.

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

**|E| = - |dV| / dr = - (potential gradient)**

Note. In the region of strong electric fields equipotential surfaces are close tighter and in the region of weak electric fields equipotential surface are far apart. This follows from

**E = - dV / dr, dr = - dV / E or dr ∝ 1/ E for given dV**.

Hence when E is large dr is mall distances between equipotential surfaces small, equipotential surfaces are crowded. The reverse is also true.

It should be clearly understood that electric potential is a scalar but electric potential gradient is a vector as it is numerically equal to electric field intensity.

A uniform electric field of** 10 N/C** exists in the vertically down wards direction. Find the increase in the electric potential as one goes up through a height of 50 cm.

**Sol: Here, E = - 10 N/C** (neg. sing for downward direction)

**dV = ? dr = 50 cm = 1/2m.**

As E = dV / dr

∴ dV = - E dr = - (-10_) x ½ = 5 volt.

The electric potential V at any point (x, y, z,) in space is given by **V = 4 x2 volt**. Calculate electric intensity at the point **(1 m, 0,2m)**.

**Sol: **Here **V = 4 x2 E = ?**

**X = 1m, y = 0, z = 2m**

E = - ∂V / ∂x = - ∂ / ∂x (4x^{2}) = - 8x = - 8 (1) = - 8 Vm^{-1}

Along negative direction of x – axis.

Electrical Potential Relation Assignment Help, Electrical Potential Relation Homework Help, Electrical Potential Relation Tutors, Electrical Potential Relation Solutions, Electrical Potential Relation Tutors, Electrostatics Help, Physics Tutors, Electrical Potential Relation Questions Answers

**Electrical Potential Relation**

**VA = V**and potential of B be

**VB = (V + dV)**where dV is increase in potential in the direction of electric intensity E normal to A and B.

Suppose dr is perpendicular distance between the two equipotential surfaces. When a unit positive charge is taken along this perpendicular distance from the surface B to the surface A against the electric field.

Work done

**WBA = | E | dr**

By definition

**WBA = VA – VB = V – (V + dV) = - dV**

∴ |E| dr = - dV

|E| = - dV / dr

∴ |E| dr = - dV

|E| = - dV / dr

Since

**|E|**is positive dV is negative.

Therefore the direction of electric field E is in the direction of decreasing potential.

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

**|E| = - |dV| / dr = - (potential gradient)**

Note. In the region of strong electric fields equipotential surfaces are close tighter and in the region of weak electric fields equipotential surface are far apart. This follows from

**E = - dV / dr, dr = - dV / E or dr ∝ 1/ E for given dV**.

Hence when E is large dr is mall distances between equipotential surfaces small, equipotential surfaces are crowded. The reverse is also true.

It should be clearly understood that electric potential is a scalar but electric potential gradient is a vector as it is numerically equal to electric field intensity.

A uniform electric field of

**10 N/C**exists in the vertically down wards direction. Find the increase in the electric potential as one goes up through a height of 50 cm.

**Sol: Here, E = - 10 N/C**(neg. sing for downward direction)

**dV = ? dr = 50 cm = 1/2m.**

As E = dV / dr

∴ dV = - E dr = - (-10_) x ½ = 5 volt.

As E = dV / dr

∴ dV = - E dr = - (-10_) x ½ = 5 volt.

The electric potential V at any point (x, y, z,) in space is given by

**V = 4 x2 volt**. Calculate electric intensity at the point

**(1 m, 0,2m)**.

**Sol:**Here

**V = 4 x2 E = ?**

**X = 1m, y = 0, z = 2m**

E = - ∂V / ∂x = - ∂ / ∂x (4x

E = - ∂V / ∂x = - ∂ / ∂x (4x

^{2}) = - 8x = - 8 (1) = - 8 Vm^{-1}

Along negative direction of x – axis.