physics

First question:
The diagram of the scenario can be shown as follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Phy/EForce1.jpg

So applying the formulae of electric force, we can find that:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Phy/EForce2.jpg

in which the negative sign of F means that it is an attractive force. And from the value of θ, we can say that:
The direction of the force on the negative charge is in the direction of 45° with the positive x-axis, in clockeise direction.
The direction of the force on the positive charge is in the direction of 135° with the positive x-axis, in anti-clockeise direction.
Q9) For (a) and (b), the value can be obtained using formula as follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Phy/EForce3.jpg

now, for (c), since the value of E-field is zero, by the inverse square law, r →∞ and so the distance is infinitely large.
Conversely, for (d), if the value of E-field is to be infinity, r→0 which implies that the position is at the point charge itself.

2007-02-14 11:11:56 補充：
Applying the formula for E=Q/4πεr^2, we have:Eo at -4uC charge = 749 V/m in the direction of towards it along the dotted line.Eo at 十6uC charge = 500 V/m in the direction away it along the dotted line.