F.5 Physics 電學 MC

1st Statement:
You may consider the electric force acting on a small +ve charge (i.e. a test charge). Just think at which point the test charge should be placed such that the force acting on it is zero.

Clearly, the point of zero net force would not be X and Y. It is because at these two points, the +ve test charge will be attracted by -Q and repelled by +2Q. Thus the forces acting on the test charge, given by +2Q and -Q, are in the same direction, there is a net force pulling the test charge to the right.

At point Z, the attractive force given by -Q is towards the left, whereas the repulsive force given by +2Q is towards the right. The two forces are in opposite directions, and would balance one another, giving a zero net force on the test charge. This indicates that the electric field (which is the force per unit charge) at Z is zero.

2nd Statement:
Electric field lines run from the charge +2Q to the charge -Q. By the properties of field lines, electric potential decreases along the lines. In other words, the point of zero potential is somewhere between the 2 charges. Because the +ve charge is two times more than the -ve charge, thus shifting the point of zero potential nearer to the -ve charge.

Mathematically, potential V at a point between the two charges is
V = k(2Q)/r1 + k(-Q)/r2
where k is the electrostatic constant, r1 and r2 are the distance of the point from the respective charges
At point of zero potential, V=0
hence, 2kQ/r1 = kQ/r2
r1/r2 = 2
r1 = 2(r2)
Thus, Y is the point of zero potential.

The answer is therefore option B.