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Question

Three charges each equal to $$+4nC$$ are placed at the three corners of a square of side $$2\ cm$$. Find the electric field at the fourth corner.


Solution

$$AD=2\ cm=2\times { 10 }^{ -2 }m$$
$$CD=2\ cm=2\times { 10 }^{ -2 }m$$
$$BD=2\sqrt { 2 }\ cm=2\sqrt { 2 } \times { 10 }^{ -2 }m$$
$${ E }_{ A }=\cfrac { 1 }{ 4\pi { \varepsilon  }_{ 0 } } \cfrac { { Q }_{  } }{ { r }^{ 2 } } =9\times { 10 }^{ 9 }\times \cfrac { 4\times { 10 }^{ -9 } }{ { \left( 2\times { 10 }^{ -2 } \right)  }^{ 2 } } =9\times { 10 }^{ 4 }N{ C }^{ -1 } $$ along $$AD$$
$${ E }_{ C }=\cfrac { 1 }{ 4\pi { \varepsilon  }_{ 0 } } =9\times { 10 }^{ 9 }\times \cfrac { 4\times { 10 }^{ -9 } }{ { \left( 2\times { 10 }^{ -2 } \right)  }^{ 2 } } =9\times { 10 }^{ 4 }N{ C }^{ -1 }$$ along $$CD$$
$${ E }_{ B }=\cfrac { 1 }{ 4\pi { \varepsilon  }_{ 0 } } \cfrac { { Q }_{ B } }{ { r }^{ 2 } } =9\times { 10 }^{ 9 }\times \cfrac { 4\times { 10 }^{ -9 } }{ { \left( 2\sqrt { 2 } \times { 10 }^{ -2 } \right)  }^{ 2 } } =4.5\times { 10 }^{ 4 }N{ C }^{ -1 }$$ along $$BD$$

$$X-$$component of the field $${ E }_{ x }={ E }_{ A }+{ E }_{ B }\cos { 45 } =9\times { 10 }^{ 4 }+4.5\times { 10 }^{ 4 }\times 0.707$$
                                              $${ E }_{ x }=12.1815\times { 10 }^{ 4 }N{ C }^{ -1 }$$

$$Y-$$component of the field $${ E }_{ y }={ E }_{ C }+{ E }_{ B }\cos { 45 } =9\times { 10 }^{ 4 }+4.5\times { 10 }^{ 4 }\times 0.707$$
                                             $${ E }_{ x }=12.1815\times { 10 }^{ 4 }N{ C }^{ -1 }$$

Magnitude $$E=\sqrt { { E }_{ x }^{ 2 }+{ E }_{ y }^{ 2 } } =\sqrt { { \left( 12.1815\times { 10 }^{ 4 } \right)  }^{ 2 }+{ \left( 12.1815\times { 10 }^{ 4 } \right)  }^{ 2 } } $$
$$E=\sqrt { 2\times { \left( 12.1815\times { 10 }^{ 4 } \right)  }^{ 2 } } =17.2272\times { 10 }^{ 4 }N{ C }^{ -1 }$$
OR
$${ E }_{ AC }=\sqrt { { E }_{ A }^{ 2 }+{ E }_{ C }^{ 2 } } =\sqrt { { \left( 9\times { 10 }^{ 4 } \right)  }^{ 2 }+{ \left( 9\times { 10 }^{ 4 } \right)  }^{ 2 } } $$
$${ E }_{ AC }=\sqrt { 2{ \left( 9\times { 10 }^{ 4 } \right)  }^{ 2 } } =12.726\times { 10 }^{ 4 }N{ C }^{ -1 }$$ along $$BD$$
$$E={ E }_{ B }+{ E }_{ AC }=4.5\times { 10 }^{ 4 }+12.724\times { 10 }^{ 4 }=17.226\times { 10 }^{ 4 }N{ C }^{ -1 }$$ along $$BD$$

874673_947103_ans_22bce4ed8060454bb456f2fbea189ed2.png

Physics
NCERT
Standard XII

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