four point charges placed at the corner of the square at a distance find the magnitude at one of the corner...
Answers
Answer:
We know that electric field lines come outward from a positive charge i.e. the electric field is in outward direction from a positive charge .All the given charges are positive therefore electric field will be outward from them .
Now the electric field E due to a point charge Q at a distance r is given by ,
E=kQ/r
2
,
the distance of any charge out of the given four , from center of square is half of the diagonal of square which will be D/
2
,
therefore electric field due to a given charge Q at center is ,
E=kQ/(D/
2
)
2
the four charges are of equal magnitude and also at equal distance from center of square so they will produce same magnitude of electric field at center but as direction of electric field by two diagonally opposite charges is opposite to each other therefore they will cancel each other and other two will do the same. Hence the net electric field at center would be zero.
Answer:
ㅛ ㅗㅐㅔㄷㅑㅅ ㅑㄴ ㅊㅐㄱㄱㄷㅊㅅ
Explanation:
Let the side of the square be =a
Potential at centre
V
c
=
2
a
K×−Q
+
2
a
K×−q
+
2
a
K×2q
+
2
a
K×2Q
0=
2
a
KQ
+
2
a
Kq
⇒Q=−q