CoursesGATEElectrical Engineering (EE) | Topic-wise Previous Solved GATE Papers | Electromagnetic Fields

- Lecture1.1
  
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- Lecture2.1
  
  Basics [Coordinate Systems/Vector Calculus]
- Lecture3.1
  
  Electrostatics
- Lecture4.1
  
  Magnetic Fields
- Lecture5.1
  
  Maxwell’s Equations and EM wave propagation

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Q.1 An electrostatic potential is given by $\theta=2x\sqrt y$ volts in the rectangular co-ordinate system. The magnitude of the electric field at x = 1 m, y = 1 m is ________ V/m. [1992 1M]

Ans:- $\sqrt5$

Explaination: – Electrostatic potential, $\theta=2x\sqrt y$

$\therefore\overrightarrow E\;=-grad\;\left(\theta\right)=-\nabla\theta$

$\therefore\overrightarrow E\;=-\left[{\widehat a}_x\;\frac\partial{\partial x}\;\left(2x\sqrt y\;\right)+{\widehat a}_y\;\frac\partial{\partial y}\;\left(2x\sqrt y\;\right)\right]\;$

$\overrightarrow E\;=\;-{\widehat a}_x\;2\sqrt y\;-{\widehat a}_{y\;}\frac x{\sqrt y}\;$

Now, at x = 1 m and y = 1 m

$\overrightarrow E\;=\;-2{\widehat a}_x\;-{\widehat a}_y$

$\therefore \left|\overrightarrow E\right|\;=\;\sqrt{2^2\;+1^2}\;=\sqrt5\;V/m$

Q.2 Given a vector field $\overrightarrow F$ , the divergence theorem states that [2002 1M]

(a) $\int_s\;\overrightarrow F.\;d\overrightarrow s\;=\;\int_v\;\nabla.\;\overrightarrow F\;dV$

(b) $\int_s\;\overrightarrow F.\;d\overrightarrow s\;=\;\int_v\;\nabla\times\;\overrightarrow F\;dV$

(d) $\int_s\;\overrightarrow F\times\;d\overrightarrow s\;=\;\int_v\;\nabla\times\overrightarrow F\;dV$

Ans:- (a)

Explanation:- The Divergence Theorem is applicable to the closed surface. It is simply remembered as, the surface integration of the vector field over the closed surface can also be calculated by taking the volume integration of the divergence of that vector field.

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