Question

Assuming that you remain a finite distance from the origin, where in the X-Y plane could a point charge Q be placed, so that this new charge experiences no electric force. Express the location in vector components, separated by a comma

Answer 1

Answer:

< 0 , -0.4494*L >

Explanation:

This problem can be fairly complex if we use vector analysis with one equation F_net = 0 and 2 unknowns x and y for the position of charge +Q.

We will simply our problem to scalar so that we are reduced down to 1 equation F_net and 1 unknown x or y.

With some intuition, we can see that +Q charge must be placed on y-axis. This  is because any position in x direction will produce a Electrostatic force component in y-direction; hence, our target to achieve F_net would be impossible. So, now we know that +Q charge must be placed on y-axis @ x= 0.

The next step is to determine "where" on the y-axis for which we have three cases.

Case 1: Above +3Q charge

If we place +Q charge above +3Q i.e y > + L, we can see that the distance between +3Q (higher in magnitude) is ( y - L ) and distance from -2Q (smaller in magnitude) is ( y ). Since, Electrostatic force is proportional to magnitude of charge and inversely related to squared of the distance. We can see that F_-2Q < F_+3Q. Hence, F_net will not be zero. (Case Rejected)

Case 2: In between +3Q and -2Q

If we place +Q charge in between +3Q & -2Q i.e 0 < y < L, we can see that the direction of Force applied by +3Q is downward and direction of Force applied by -2Q is also downwards. We can see that F_net = F_-2Q + F_+3Q. Hence, F_net will not be zero. (Case Rejected)

Case 3: Below -2Q charge

If we place +Q charge below -2Q i.e y < 0, we can see that the distance between +3Q (higher in magnitude) is ( y + L ) and distance from -2Q (smaller in magnitude) is ( L ). Since, Electrostatic force is proportional to magnitude of charge and inversely related to squared of the distance. We can see that F_-2Q may be equal to F_+3Q at distance y. Hence, F_net will may be zero. (Case Accepted)

Using Case 3:

Below -2Q charge @ ( 0 , y )

F_+3Q = k*(+Q)*(+3Q) / (y+L)^2

F_-2Q =  k*(+Q)*(-2Q) / (y)^2

F_net = F_+3Q - F_-2Q = 0

(3 / (y+L)^2 ) - (2 / y^2) = 0

3y^2 - 2(y + L )^2 = 0

y^2 - 4Ly - 2L^2 = 0

Using quadratic formula:

y = [2 + sqrt(6)] * L = 4.4494*L

y = [2 - sqrt(6)] * L = -0.4494*L (Accepted)

Hence,

we determined from case 3 above that y < 0, so our position vector for +Q charge will be < 0 , -0.4494*L >