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GATE EE 2018 Official Paper

Option 4 : -999.09 V/m

CT 1: Ratio and Proportion

3742

10 Questions
16 Marks
30 Mins

Given that a positive charge of 1 nc is placed at (0, 0, 0.2).

According to method image charge concept, there will be an image charge below the conducting ground plane.

Electric field is given by

\(\bar E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{Q\bar R}}{{{{\left| {\bar R} \right|}^3}}}\)

Electric field at point P due to point charge + 1 nc,

\(\bar E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{1 \times {{10}^{ - 9}}\left[ {\left( {0 - 0} \right){{\bar a}_x} + \left( {0 - 0} \right){{\bar a}_y} + \left( {0.1 - 0.2} \right){{\bar a}_z}} \right]}}{{{{\left( {0.1} \right)}^3}}}\)

\(= \frac{1}{{4\pi \times 8.85 \times {{10}^{ - 12}}}} \times \frac{{1 \times {{10}^{ - 9}}}}{{{{\left( {0.1} \right)}^3}}}\left( { - 0.1} \right){\bar a_z}\)

= -899.18 V/m

Electric field at point P, due to point charge -1 nc,

\({\bar E_2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{ - 1 \times {{10}^{ - 9}}\left[ {\left( {0 - 0} \right){{\bar a}_x} + \left( {0 - 0} \right){{\bar a}_y} + \left( {0.1 + 0.2} \right){{\bar a}_z}} \right]}}{{{{\left( {0.3} \right)}^3}}}\)

\( = \frac{1}{{4\pi \times 0.85 \times {{10}^{ - 12}}}} \times \frac{{\left( { - 1 \times {{10}^{ - 9}}} \right)}}{{{{\left( {0.3} \right)}^3}}}\left( {0.3} \right){\bar a_z} = - 99.90\;V/m\)

E̅ = E̅_{1} + E̅_{2} = -899.18 V/m – 99.90 V/m