Poisson's and Laplace's Equation MCQ Quiz - Objective Question with Answer for Poisson's and Laplace's Equation - Download Free PDF

Explanation:   

The potential V(r, θ) is expressed as:
   
   \(V(r, θ) = \sum_{n=0}^{\infty} \left( A_n r^n + B_n \frac{1}{r^{n+1}} \right) P_n(\cos θ)\)
   
   Where:
    \(A_n\) and \(B_n\) are coefficients related to the potential.
    \(P_n\) are the associated Legendre polynomials.
   The potential at a specific radius R can be simplified by substituting values into the potential expression:
   
   \(V = A_r R^0 + \frac{B_0}{R}\)
   
   Thus:
   
   \(A_0 = V - \frac{B_0}{R} = V + 2V = 3V\)
   
   The relation can be manipulated:
   
   \(2V = A_0(2R^0) + \frac{B_0}{2R}\)
   
   Which simplifies to:
   
   \(V = \frac{B_0}{2R} - \frac{B_0}{2R}\)
   
   Continuing the evaluation:
   
   \(V = -\frac{B_0}{2} \quad \Rightarrow \quad B_0 = -2RV\)
   
   The potential at\( V(\frac{3}{2}R)\) is given as:
   
 \( V\left( \frac{3}{2} R \right) = A_0 + \frac{B_0}{\frac{3}{2} R} = 3V + \frac{2}{3R} - 2 = 3V - 2\)
   
   The final expression becomes:
   
   \(\left( \frac{3 - 4}{3} \right)V = \frac{5}{3}V \quad \Rightarrow \quad f = 1.66\)
   The value of f is 1.66.

Concept:

In the cartesian coordinate system, the generalized heat conduction equation is given by 

\(\frac{{{\partial ^2}T}}{{\partial {x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} + \frac{{{\partial ^2}T}}{{\partial {z^2}}} + \frac{q}{k} =\frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}\)      ....(1)

At a steady state without internal heat generation and constant thermal conductivity, the conduction equation becomes 

\(\frac{{{\partial ^2}T}}{{\partial {x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} + \frac{{{\partial ^2}T}}{{\partial {z^2}}} = 0\)

∇2T = 0 ⇒ Laplace equation 

At a steady state with internal heat generation and constant thermal conductivity, the conduction equation becomes 

\({\nabla ^2}T + \frac{q}{k} = 0 ⇒\)Poisson's equation

At an unsteady state without internal heat generation and constant thermal conductivity, the conduction equation becomes 

\({{\rm{\nabla }}^2}T = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}\) ⇒ Diffusion Equation

Explanation:

Let u(x, y, t) = Temperature of the plate at the position in 2 D

In the steady state of a two-dimensional heat flow, If u is independent of t, then the equation (1) reduces to,

\(\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0\)

⇒ Laplace’s equation in two dimensions

Poisson’s equation is derived from Columb’s law and Gauss’s theorem.

The Poisson’s equation for electrostatics is given by:

\({\nabla ^2}V = - \frac{\rho }{\varepsilon }\)

∇ = Divergence operation

ε = Permittivity of medium

ρ = charge density

For a given charge density ‘ρ’, the potential function can be obtained from the above equation.

Note: If charge density is zero, then it results in Laplace equation

Poisson's and Laplace equation:

Poisson's equation is:

\(∇^2V = -{ρ_v \over \epsilon}\)

For charge free area, ρv = 0:

\(∇^2V = 0\)

This is Laplace equation.

Additional Information Kirchhoff's Voltage Law (KVL):

The sum of voltage drops in a closed loop is equal to zero.

V1 + V₂ – IR1 – IR2 = 0

Or, V1 + V₂ = IR1 + IR2

ΣV = 0

Maxwell Equations:

Poisson's and Laplace equation:

Poisson's equation is:

\(\nabla^2V = -{ρ_v \over \epsilon}\)

For charge free area, ρ_v = 0:

\(\nabla^2V = 0\)

This is Laplace equation.

Laplace equation in the cylindrical coordinate system is:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)\(\frac{1}{\rho} \frac{\partial}{\partial \rho}\left(\rho \frac{\partial V}{\partial \rho}\right)+\frac{1}{\rho^{2}} \frac{\partial^{2} V}{\partial \phi^{2}}+{{\partial ^2V\over \partial z^2}}=0\)

if z = 0, then 

\(\frac{1}{\rho} \frac{\partial}{\partial \rho}\left(\rho \frac{\partial V}{\partial \rho}\right)+\frac{1}{\rho^{2}} \frac{\partial^{2} V}{\partial \phi^{2}}=0\)

This is Laplace's equation with no Z dependence.

Concept:

The capacitance is given by the formula:

\({C}= \frac{{A\epsilon}}{W}\)

Analysis:

The two capacitors developed are in parallel. Hence the breakdown voltage of the combined capacitor will remain V.

Q_eq = C_eqV

\({C_{eq}} = \frac{{{\varepsilon _0}A}}{{2d}} + \frac{{6{\varepsilon _0}A}}{{2d}} = 3.5\;C\)

Q_eq = 3.5 C × V

Since the total charge stored Q will be C × V, we can write:

Qeq = 3.5 Q

Gauss Law:

This law gives the relation between the distribution of electric charge and the resulting electric field.

According to this law, the total charge Q enclosed in a closed surface is proportional to the total flux ϕ enclosed by the surface, i.e.

ϕ ∝ Q

The Gauss law formula is expressed by:

\(\phi =\frac{Q}{ϵ_0}\)

This is explained with the help of the following figure:

Calculation:

From  Gauss Law:

∇.D̅  = ρv

As D̅ = ϵE̅ 

So ∇.(ϵE̅) = ρv        ----(1)

Since the medium is homogenous and isotropic,

\(So \ \nabla.\bar E=\frac{\rho_v}{\epsilon}\)  

\(Also, \ \bar E=-\frac{dv}{dr}=-\nabla V\)

Putting the value of E in equation (1)

\(\nabla .(-\nabla V)=\frac{\rho_v}{\epsilon}\)

\(\nabla^2V=-\frac{\rho_v}{\epsilon}\)

Poisson’s equation is derived from Columb’s law and Gauss’s theorem.

The Poisson’s equation for electrostatics is given by:

\({\nabla ^2}V = - \frac{\rho }{\varepsilon }\)

∇ = Divergence operation

ε = Permittivity of medium

ρ = charge density

For a given charge density ‘ρ’, the potential function can be obtained from the above equation.

Note: If charge density is zero, then it results in Laplace equation

Poisson's and Laplace equation:

Poisson's equation is:

\(∇^2V = -{ρ_v \over \epsilon}\)

For charge free area, ρv = 0:

\(∇^2V = 0\)

This is Laplace equation.

Additional Information Kirchhoff's Voltage Law (KVL):

The sum of voltage drops in a closed loop is equal to zero.

V1 + V₂ – IR1 – IR2 = 0

Or, V1 + V₂ = IR1 + IR2

ΣV = 0

Maxwell Equations:

Explanation: 

Laplace’s equation states that the sum of the second-order partial derivatives of U (function) with respect to the coordinates, equals zero.

∇2 is called the Laplacian or Laplace operator.

In Cartesian coordinates:

\({∇ ^2}U = \frac{{{\partial ^2}U}}{{\partial {x^2}}} + \frac{{{\partial ^2}U}}{{\partial {y^2}}} + \frac{{{\partial ^2}U}}{{\partial {z^2}}} = 0\)

In Cylindrical coordinates:

\({∇ ^2}U = \frac{1}{ρ}\frac{\partial }{{\partial ρ}}\left( {ρ\frac{{\partial U}}{{\partial ρ}}} \right) + \frac{1}{{{ρ^2}}}\frac{{{\partial ^2}U}}{{\partial {{\rm{\Theta }}^2}}} + \frac{{{\partial ^2}U}}{{\partial {z^2}}} = 0\)

Analysis:

V = A log ρ + B

\({∇ ^2}V = \frac{1}{ρ}\frac{\partial }{{\partial ρ}}\left( {ρ\frac{{\partial V}}{{\partial ρ}}} \right) + \frac{1}{{{ρ^2}}}\frac{{{\partial ^2}V}}{{\partial {{\rm{\Theta }}^2}}} + \frac{{{\partial ^2}V}}{{\partial {z^2}}} \)

\({∇ ^2}V = \frac{A}{ρ}\frac{\partial }{{\partial ρ}}\left( {ρ\frac{{1}}{{ ρ}}} \right) + 0 + 0\)

= 0

∴ It satisfies the Laplace equation. 

Concept:

Gauss Law:

This law gives the relation between the distribution of electric charge and the resulting electric field.

According to this law, the total charge Q enclosed in a closed surface is proportional to the total flux ϕ enclosed by the surface, i.e.

ϕ ∝ Q

The Gauss law formula is expressed by:

\(\phi =\frac{Q}{ϵ_0}\)

This is explained with the help of the following figure:

Calculation:

From  Gauss Law:

∇.D̅  = ρ_v

As D̅ = ϵE̅ 

So ∇.(ϵE̅) = ρ_v  ---(1)

Since the medium is homogenous and isotropic,

\(So \ \nabla.\bar E=\frac{\rho_v}{\epsilon}\)  

\(Also, \ \bar E=-\frac{dv}{dr}=-\nabla V\)

Putting the value of E in equation (1)

\(\nabla .(-\nabla V)=\frac{\rho_v}{\epsilon}\)

\(\nabla^2V=-\frac{\rho_v}{\epsilon}\)

Hence option(2) is the correct answer.

Poisson's and Laplace equation:

Poisson's equation is:

\(\nabla^2V = -{ρ_v \over \epsilon}\)

For charge free area, ρ_v = 0:

\(\nabla^2V = 0\)

This is Laplace equation.

Laplace equation in the cylindrical coordinate system is:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)\(\frac{1}{\rho} \frac{\partial}{\partial \rho}\left(\rho \frac{\partial V}{\partial \rho}\right)+\frac{1}{\rho^{2}} \frac{\partial^{2} V}{\partial \phi^{2}}+{{\partial ^2V\over \partial z^2}}=0\)

if z = 0, then 

\(\frac{1}{\rho} \frac{\partial}{\partial \rho}\left(\rho \frac{\partial V}{\partial \rho}\right)+\frac{1}{\rho^{2}} \frac{\partial^{2} V}{\partial \phi^{2}}=0\)

This is Laplace's equation with no Z dependence.

Concept:

The condition for potential (V) to satisfy Laplace's equation is given by:

• \(\nabla^2V = 0\)

Laplace equations in different coordinate systems are given by:

• Cartesian System: \(\nabla^2V = { \partial^2V\over \partial x^2}+{ \partial^2V\over \partial y^2}+{ \partial^2V\over \partial z^2}\)
• Cylindrical System: \(\nabla^2V ={1\overρ}{\partial \over \partialρ }(ρ \space {\partial V\over \partialρ})+{{1\over ρ^2}{ \partial^2V\over \partial ϕ^2}}+{ \partial^2V\over \partial z^2}\)
• Spherical System: \(\nabla^2V ={1\over r^2}{\partial \over \partial r }(r^2 \space {\partial V\over \partial r})+{1\over r^2s in\theta}{\partial \over \partial \theta }(sin\theta\space {\partial V\over \partial \theta})+{1\over r^2sin^2\theta }{ \partial^2V\over \partial ϕ^2}\)

​Calculation:

Option 1: V = 2x + 5

\(\nabla^2V = { \partial^2(2x+5)\over \partial x^2}+0\)

\(\nabla^2V = 0\)

It satisfies the Laplace equation.

Option 2: \(V = \frac{{10}}{ρ}\)

 \(\nabla^2V ={1\overρ}{\partial \over \partialρ }(ρ \space {\partial \over \partialρ}(\frac{{10}}{ρ}))+0\)

\(\nabla^2V ={1\overρ}{\partial \over \partialρ }(ρ (\frac{{-10}}{ρ^2}))\)

\(\nabla^2V ={1\overρ}{\partial \over \partialρ } (\frac{{-10}}{ρ})\)

\(\nabla^2V =\frac{{10}}{ρ^3} \neq 0\)

It does not satisfy the Laplace equation.

Option 3: V = 10xy

\(\nabla^2V = { \partial^2(10xy)\over \partial x^2}+{ \partial^2(10xy)\over \partial y^2}+0\)

\(\nabla^2V = 0\)

It satisfies the Laplace equation.

Option 4: V = ρ cosϕ 

\(\nabla^2V ={1\overρ}{\partial \over \partialρ }(ρ \space {\partial (\rho cos\phi)\over \partialρ})+{{1\over ρ^2}{ \partial^2(\rho cos\phi)\over \partial ϕ^2}}+0\)

\(\nabla^2V = 0\)

It satisfies the Laplace equation.

Properties of Laplace equation:

∇²f = 0 (Laplace equation)

1) The solutions have neither maxima nor minima anywhere except at the boundaries.

2) The solution is separable in co-ordinates

∇²f = 0

\(\frac{{{\partial ^2}{f_x}}}{{\partial {x^2}}} + \frac{{{\partial ^2}{f_y}}}{{\partial {y^2}}} + \frac{{{\partial ^2}{f_z}}}{{\partial {z^2}}} = 0\) 

Where f = f_x a_x + f_y a_y + f_z a_z

3) The solutions are continuous.

4) The solutions are dependent on the boundary conditions.

So, only option A is correct.   

Concept:

The capacitance of a parallel plate capacitor is given by

\(C = \frac{{\varepsilon A}}{d} = \frac{{{\varepsilon _0}{\varepsilon _r}A}}{d}\)

Where ε_r is the relative permittivity

ε₀ is the absolute permittivity

A = area of plates

d = distance between the two plates.

Calculation:

Rolled paper capacitor C = 0.02 μF

Width of Aluminium strips (w) = 6 cm = 0.06 m

Rolled paper capacitor means an Aluminium strip is rolled with parallel capacitor plates.

Therefore, the area of that parallel plate = Area of Al

Area of Al. (A) = ω × ℓ = 0.06 × ℓ m²

Distance between two parallel plates = thickness of Al. strips

= 0.06 mm = 0.06 × 10^-3 m

Now, \(C = \frac{{{\varepsilon _r}{\varepsilon _0}A}}{d} = \frac{{3 \times 8.85 \times {{10}^{ - 12}} \times .06 \times \ell }}{{0.06 \times {{10}^{ - 3}}}}\)

⇒ 0.02 × 10^-6 = 3 × 8.85 × 10^-9 × ℓ

⇒ ℓ = 0.7765 m