Question

Three charges q, −q and q₀ are placed as shown in figure. The magnitude of the net force on the charge q₀ at point O is `[k = 1/((4 pi epsilon_0))]`

पर्याय

• 0

• `(2kqq_0)/a^2`

• `(sqrt 2 kqq_0)/a^2`

• `1/sqrt 2 (kqq_0)/a^2`

Answer 1

`bb((sqrt 2 kqq_0)/a^2)`

Explanation:

Positions of q₀, −q and q are shown

Both −q and q charges are equidistant from q₀.

So, the magnitude of both forces on q₀ will be equal. The angle between the forces will be 90° as shown in the diagram.

Hence the resultant force

= `sqrt (F^2 + F^2 + 2F xx F xx cos 90^circ)`

= `sqrt 2 F`

= `sqrt 2 xx (kqq_0)/a^2`

= `sqrt 2 xx (1/(4 pi epsilon_0)) xx ((qq_0)/a^2)`