1.

The corners A, B, C, and D of a square are occupied by charges q,-q,2Q, and Q, respectively. The side of square is 2b. The field at the midpoint of side CD is zero. What is the value of q//Q?

Answer»

`5sart5//2`
`2sqrt2//5`
`2//5`
`5//2`

Solution :a. Since the field at midpoint of `CD` is ZERO. Hence. `EB cos+ED+Eacos-EC=0`
`((q)/(4piepsilon_(0)(sqrt(5B))^(2))+(q)/(4piepsilon_(0)(sqrt(5b))^(2)))`
or `costheta+(Q)/(4piepsilon_(0)b^(2))=(2Q)/(4piepsilon_(0)b^(2))`
On SOLVING, we get
`(2q)/(5)costheta=Q` or `(2q)/(5)((b)/(sqrt(5b)))=Q`
or `(2q)/(5sqrt(5))=Q` or `(q)/(Q)=(5sqrt(5))/(2)`


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