Verified Solution • 2025–26 Session moderate

QJEEM19Phfe44: JEE Main 2019 Physics - Class 12 JEEM19Phfe44 Solution: Two point charges q 1 and q 2 ( 25 C) are placed on the x-axis…

Two point charges q1 (10μC)\left( {\sqrt {10} \mu C} \right) and q2( −- 25 μ\mu C) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,
[take 14π∈0{1 \over {4\pi { \in _0}}} = 9 ×\times 109 Nm2C −- 2]

Multiple Choice Options

  • A (63i^−27j^)×102\left( {63\widehat i - 27\widehat j} \right) \times {10^2}
  • B (−63i^+27j^)×102\left( { - 63\widehat i + 27\widehat j} \right) \times {10^2}
  • C (81i^−81j^)×102\left( {81\widehat i - 81\widehat j} \right) \times {10^2}
  • D (−81i^+81j^)×102\left( { - 81\widehat i + 81\widehat j} \right) \times {10^2}

Step-by-Step Verified Solution

JEE Main 2019 (Online) 9th January Evening Slot Physics - Electrostatics Question 186 English Explanation 1
Electric field due to 10 μC\sqrt {10} \,\mu C charge :
JEE Main 2019 (Online) 9th January Evening Slot Physics - Electrostatics Question 186 English Explanation 2
E1→=−\overrightarrow {{E_1}} = - E1 sin θ\theta 1 i^\widehat i + E1 cos θ\theta 1 j^\widehat j

Where,

E1 =14πε0×∣q1∣r12= {1 \over {4\pi {\varepsilon _0}}} \times {{\left| {{q_1}} \right|} \over {{r_1}^2}}

=9×109×10×10−6(12+32)2= 9 \times {10^9} \times {{\sqrt {10} \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{1^2}} + {3^2}} \right)}^2}}}

=9×10310 v/m= {{9 \times {{10}^3}} \over {\sqrt {10} }}\,v/m

sin θ\theta 1 == 110{1 \over {\sqrt {10} }}

and cos θ\theta 1 = 310{3 \over {\sqrt {10} }}

∴\therefore     E1→=9×10310 (−110i^+310j^)\overrightarrow {{E_1}} = {{9 \times {{10}^3}} \over {\sqrt {10} }}\,\left( { - {1 \over {10}}\widehat i + {3 \over {\sqrt {10} }}\widehat j} \right)

=9×102(−i^+3j^)= 9 \times {10^2}\left( { - \widehat i + 3\widehat j} \right)

Electric field due to −- 25 μ\mu C charge,

JEE Main 2019 (Online) 9th January Evening Slot Physics - Electrostatics Question 186 English Explanation 3

E2→=\overrightarrow {{E_2}} = E2 sin θ\theta 2 i^\widehat i −- E2 cos θ\theta 2 j^\widehat j

where

E2 =14πε0×∣92∣r22= {1 \over {4\pi {\varepsilon _0}}} \times {{\left| {{9_2}} \right|} \over {r_2^2}}

=9×109×25×10−6(42+32)2= 9 \times {10^9} \times {{25 \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{4^2} + {3^2}} } \right)}^2}}}

=9×103= 9 \times {10^3} V/m

sin θ\theta 2 = 45{4 \over 5}

and cos θ\theta 2 = 35{3 \over 5}

∴\therefore     E2→=9×103  (45i^−35j^)\overrightarrow {{E_2}} = 9 \times {10^3}\,\,\left( {{4 \over 5}\widehat i - {3 \over 5}\widehat j} \right)

=18×102(4i^−3j^)= 18 \times {10^2}\left( {4\widehat i - 3\widehat j} \right)

∴\therefore    Net electric field,

E→\overrightarrow E = E→1{\overrightarrow E _1} + E→2{\overrightarrow E _2}

=(63i^−27j^)×102  V/m= \left( {63\widehat i - 27\widehat j} \right) \times {10^2}\,\,V/m

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