Solved Exercise/Numericals - Electric Potential, Potential Gradient & Electric Dipole.

Q.1 Determine the electric field due to the following potential:
a) V = x² + 2y² + 4z²
Ans:

= - (2x a_x + 4y a_y + 8z a_z )
= -2x a_x - 4y a_y - 8z a_z V/m

b) V = ρ² (z + 1) sinφ

= - [ 2ρ (z + 1)sinφ a_ρ + ρ(z + 1) cosφ a_φ + ρ² sinφ a_z ]
= - 2ρ (z + 1)sinφ a_ρ - ρ(z + 1) cosφ a_φ - ρ² sinφ a_z



Q.2 A point charge of 5 nC is located at the origin. If V = 2v at (0, 6, -8), find
a) The potential at A (-3, 2, 6)
b) The potential at B (1, 5, 7)
c) The potential difference V_AB
Ans:
If V (0, 6, -8) = 2 V
In this case
x = 0; y = 6; z = -8

Using the scalar relationship between Cartesian and spherical system, we have

r = (x² + y² +z²)^1/2 = (100)^1/2 =10

Hence

Electric potential at point r due to a point charge Q located at origin

→ C = - 2.5

a) Electric potential at point r due to a point charge Q located at a point (-3, 2, 6) is given as:

= 3. 929 V

b) Electric potential at point r due to a point charge Q located at a point (1, 5, 7) is given as:

= 2.696 V

c) The potential difference V_AB is given as:

V_AB = V_B – V_A = 2.696 – 3.929 = - 1.233 V



Q.3 If point charge 3 μC is located at the origin. Also there are two more charges -4 μC and 5 μC are located at (2, -1, 3) and (0, 4, -2) respectively. Find potential at (-1, 5, 2) ? Assume zero potential at infinity.
Ans:
For N point charges Q₁, Q₂ ….Q_n located at points with position vectors r₁, r₂, r₃…..r_n, the electric potential at point r is given as:

At V ( ∞ ) = 0, C = 0

| r – r₁ | = | (-1, 5, 2) – (2, -1, 3) | = (46)^1/2
| r – r₂ | = | (-1, 5, 2) – (0, 4, -2) | = (18)^1/2
| r – r₃ | = | (-1, 5, 2) – (0, 0, 0) | = (30)^1/2

Hence electric potential is given as:

= 10.3 kV



Q.4 An electric dipole of 100 a_z pC.m is located at the origin. Find V and E at points
a) (0, 0, 10)
b) (1, π/3, π/2)
Ans:
a) At point (0, 0, 10)
Cartesian → Spherical
(x, y, z) → (r, θ, φ)

(0, 0, 10) → (10, 0^o, 0^o)

Electric potential (V) due to a electric dipole centered at origin and aligned with the z axis is written as:

Electric field intensity (E) is the negative gradient of Electric Potential (V).

b)At point (1, π/3, π/2)

Electric potential (V) due to a electric dipole centered at origin and aligned with the z axis is written as:

Electric field intensity (E) is the negative gradient of Electric Potential (V).

ALSO READ: 

- Gauss's Law - Theory.

- Gauss's Law - Application To a Point charge.

- Gauss's Law - Application To An Infinite Line Charge.

- Gauss's Law - Application To An Infinite Sheet Charge.

- Gauss's Law - Application To a Uniformly Charged Sphere.

- Numericals / Solved Examples - Gauss's Law.

- Scalar Electric Potential / Electrostatic Potential (V).

- Relationship Between Electric Field Intensity (E) and Electrostatic Potential (V).

- Electric Potential Due To a Circular Disk.

- Electric Dipole.

- Numericals / Solved Examples - Electric Potential and Electric Dipole.

- Energy Density In Electrostatic Field / Work Done To Assemble Charges.

- Numericals / Solved Examples - Electrostatic Energy and Energy Density.

- Numericals / Solved Examples - Gauss's law...



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