Electromagnetic Fields Test 1

The new drop has twice the volume of the smaller drop.

\({V_1} = \frac{4}{3}\pi {R^3}\) 

Let the radius of the larger drop is R₂

The volume of the larger drop, \({V_2} = \frac{4}{3}\pi R_2^3\) 

V₂ = 2 V₁

\( \Rightarrow \frac{4}{3}\pi R_2^3 = 2 \times \frac{4}{3}\pi {R^3}\) 

⇒ R₂ = 2^1/3 R = 1.26 R

Capacitance = 4πε₀ R₂ = 5.04 π ε₀ R

New capacitance = 1.26 C

Charge density ρ_v = 0.2 μ c/m³

r₁ = 4 cm, r₂ = 5 cm

Total charge \(Q = \mathop \smallint \nolimits_{Vol} {\rho _v}dv = \mathop \smallint \limits_{r = 0.04}^{0.05} \mathop \smallint \limits_{\theta = 0}^\pi \mathop \smallint \limits_{\phi = 0}^{2\pi } \left( {0.2 \times {{10}^{ - 6}}{r^2}\sin \theta } \right)drd\theta d\phi \) 

\( = \left( {0.2} \right)\left( {2\pi } \right) \times \left( {\frac{{{r^3}}}{3}} \right)_{0.04}^{0.05}\left( { - \cos \theta } \right)_{\theta = 0}^\pi \) 

\(= \frac{{0.4\pi }}{3}\left( {{{0.05}^3} - {{0.04}^3}} \right)\left( 2 \right) \times {10^{ - 6}}\) 

Q = 51.103 pC

The capacitance of a parallel plate capacitor is

\({C_1} = \frac{{{\varepsilon _0}A}}{d}\) 

When the distance of separation becomes 2d,

\({C_2} = \frac{{{\varepsilon _0}A}}{{2d}}\) 

⇒ C₁ = 2C₂

 Q = CV

C₁V₁ = C₂V₂

⇒ 2 C₂ V = C₂ V₂

⇒ V₂ = 2V

Initial stored energy, \({u_i} = \frac{1}{2}\;{C_1}{V^2}\) 

\(= \frac{1}{2} \times \frac{{{\varepsilon _0}A}}{d}{V^2} = \frac{{{\varepsilon _0}A}}{{2d}}{V^2}\) 

Final stored energy, \({u_f} = \frac{1}{2}{C_2}{V^2}\) 

\(= \frac{1}{2} \times \frac{{{\varepsilon _0}A}}{{2d}}{\left( {2V} \right)^2} = \frac{{{\varepsilon _0}A}}{d}{V^2}\) 

The work required to separate the plates is

\({\rm{\Delta }}W = {u_f} - {u_t} = \frac{{{\varepsilon _0}A}}{d}{V^2} - \frac{{{\varepsilon _0}A}}{{2d}}{V^2}\) 

\(= \frac{{{\varepsilon _0}A}}{{2d}}{V^2}\) 

If there were no dielectric the field between the plates would be given by Gauss’s law as

\({E_0} = \frac{q}{{A{\varepsilon _0}}}\) 

\( = \frac{{890\; \times \;{{10}^{ - 9}}}}{{100\; \times \;{{10}^{ - 4}}\; \times\; 8.85\; \times \;{{10}^{ - 12}}}} = 1.01 \times {10^7}V/m\) 

The dielectric constant is,

\(k = \frac{{{E_0}}}{E} = \frac{{1.01\; \times \;{{10}^7}}}{{1.4\; \times \;{{10}^6}}} = 7.2\) 

Total charge, q_tot = ε₀ E A

= 8.85 × 10^-12 × 1.4 × 10⁶ × 10^-2

= 124 nC

q_tot = q – q¹

Induced charge, Q’ = Q - Q_tot

= 890 – 124 = 766 nC

Concept:

In cylindrical coordinates; for a given vector \(\vec A\)

\(\vec A = {A_p}\widehat {{a_\rho }} + {A_\phi }\widehat {{a_\phi }} + {A_z}\widehat {{a_z}}\)

\(\nabla \times \vec A = \frac{1}{\rho }\left| {\begin{array}{*{20}{c}} {\widehat {{a_\rho }}}&{P\widehat {{a_\phi }}}&{\widehat {{a_z}}}\\ {\frac{\partial }{{\partial \rho }}}&{\frac{\partial }{{\partial \phi }}}&{\frac{\partial }{{\partial z}}}\\ {{A_\rho }}&{\rho {A_\phi }}&{{A_z}} \end{array}} \right|\)

Calculation:

\(\vec A = 2\rho \cos \phi \widehat {{a_z}}\)

\(\nabla \times \vec A = \frac{1}{\rho }\left| {\begin{array}{*{20}{c}} {\widehat {{a_\rho }}}&{\rho \widehat {{a_\phi }}}&{\widehat {{a_z}}}\\ {\frac{\partial }{{\partial \rho }}}&{\frac{\partial }{{\partial \phi }}}&{\frac{\partial }{{\partial z}}}\\ 0&{\rho .0}&{2\rho \cos \phi } \end{array}} \right|\)

\(= \frac{1}{\rho }\left[ {\widehat {{a_\rho }}\left( {\frac{\partial }{{\partial \phi }} \cdot 2\rho \cos \phi } \right) - \rho \widehat {{a_\phi }}\left( {\frac{\partial }{{\partial \rho }} \cdot 2\rho \cos \phi } \right)} \right]\)

\( = \frac{1}{\rho }\left[ { - 2\rho \sin \phi \widehat {{a_\rho }} - 2\rho \cos \phi \widehat {{a_\phi }}} \right] = \left( { - 2\sin \phi } \right)\widehat {{a_\rho }} - \left( {2\cos \phi } \right)\widehat {{a_\phi }}\)

\(\nabla \times \vec A\left( {at\;\left( {4,\pi ,0} \right)} \right) = - 2\left( {\sin \pi } \right)\widehat {{a_\rho }} - 2\cos \left( \pi \right)\widehat {{a_\phi }} = 2\widehat {{a_\phi }}\)