Two point charges `+5xx10^(-19)C and +20xx10^(-19)C` are separated by a distance of 2m. The electric field intensity will be zero at a distance d=………. From the charges of `5xx10^(19)C`.

Options

A. `x=2//3 m`
B. `x=-2m`,
C. `x= -12 m` or `x=2//3 m,` if `x=-2m`, the point P lies to the left of A.`
D. or `x=-2//3 m,`

Answer

B

Explanation

Here `q_(1)=+5xx10^(-19)C, q_(2)=+20xx10^(-19)C`. If P is the point (distance x form `q_(1)`), where electric field intesity is zero, then,
`|E_(1)|=|E_(2)|`
or, `K_(e)(q_(1))/(x^(2))=k_(e)(q_(2))/(2-x^(2))`
or, `(2-x^(2))/(x^(2))=(q_(2))/(q_(1))=(20xx10^(-19)C)/(5xx10^(-19)C) =4`
or , `(2-x)/(x)= +-2`
Hence, `x= -2 m` or `x=2//3 m,` if `x=-2m`, the point P lies to the left of A. In this case, since `vecE_(1)` and `vecE_(2)` are in the sme direction. the two positive charges produce a resultant electric field `vec(E)` where `vec(E)=vecE_(1)+vecE_(2)` as shown in fig. Hence, this is not a feasible situation.
,
Thus, as shown in fig. `x=2//3m`.