This question was previously asked in

JEE Mains Previous Paper 1 (Held On: 10 Jan 2019 Shift 2)

Option 4 : \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\left( {1 + \frac{1}{{\sqrt 5 }}} \right)\)

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

8346

90 Questions
360 Marks
180 Mins

**Concept:**

Electric potential at origin (0, 0) due to these charges can be found by scalar addition of electric potentials due to each charge.

\(\therefore V = \frac{{KQ}}{{{r_1}}} + \frac{{KQ}}{{{r_2}}} + \frac{{KQ}}{{{r_3}}} + \frac{{KQ}}{{{r_4}}}\)

If another Q charge is placed at origin, then work done to get the charge at origin

W = QV

**Calculation:**

Given,

The four charges are shown in the figure below

Electric potential at origin (0, 0) due to these charges can be found by scalar addition of electric potentials due to each charge.

\(\therefore V = \frac{{KQ}}{{{r_1}}} + \frac{{KQ}}{{{r_2}}} + \frac{{KQ}}{{{r_3}}} + \frac{{KQ}}{{{r_4}}}\) ----(1)

Here,

r_{1} = 2, r_{2} = 2

\({r_3} = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( {2 - 0} \right)}^2}} = \sqrt {{4^2} + {2^2}} = \sqrt {16 + 4} = \sqrt {20} \)

\({r_4} = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( { - 2 - 0} \right)}^2}} = \sqrt {{4^2} + {2^2}} = \sqrt {16 + 4} = \sqrt {20} \)

Substituting these values in the equation (1),

\(V = KQ\left[ {\frac{1}{2} + \frac{1}{2} + \frac{1}{{\sqrt {20} }} + \frac{1}{{\sqrt {20} }}} \right]\)

\(= KQ\left[ {1 + \frac{2}{{\sqrt {20} }}} \right]\)

\(= KQ\left[ {1 + \frac{1}{{\sqrt 5 }}} \right]\)

\(V = \frac{{KQ\left( {\sqrt 5 + 1} \right)}}{{\sqrt 5 }}\;volt\) ----(2)

Now, if another Q charge is placed at origin, then work done to get the charge at origin

\(W = QV\) ----(3)

By putting value of V from equation (2) in equation (3), we get

\(W = K{Q^2}\frac{{\left( {\sqrt 5 + 1} \right)}}{{\sqrt 5 }}\;J\)

We know that, \(K = \frac{1}{{4\pi {\epsilon_0}}}\)

\(W = \frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\left( {1 + \frac{1}{{\sqrt 5 }}} \right)\;J\)

The work required to put a fifth charge Q at the origin of the coordinate system will be \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\left( {1 + \frac{1}{{\sqrt 5 }}} \right)\;J\).