Two point charges q_{1 }( ) and q_{2}(−25 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,

a. |
(−63 ) × 10 |
b. |
(81 -81 |

c. |
(63 × 10 |
d. |
(−81 ×10 |

Two point charges q_{1 }( ) and q_{2}(−25 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,

a. |
(−63 ) × 10 |
b. |
(81 -81 |

c. |
(63 × 10 |
d. |
(−81 ×10 |

1 Answer

127 votes

Let & are the vaues of electric field due to q_{1 } and q_{2 }Respectively magnitude of

E_{2 = } _{ }

E_{2= }

E_{2}=9×10^{3 }V/m

∴ ^{ } ^{ }= 9×10^{3 }( − sinθ_{2 }

tanθ_{2 }= 3/4

∴ ^{ } ^{ }= 9×10^{3 }

Magnitude of =

=(9

=9 10^{2}

∴ ^{ } =9 10^{2} [cos _{1}( )+sin _{1} ]

∴ tan =3

E_{1}=9 10^{2}

E_{1}=9 10^{2} = 10^{2}

∴ = _{1}+ _{2 }=(63 −27 ) 10^{2} V/m

∴correct answer is (c)

127 votes

127