Identical +7.74 ƒÊC charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total potential at the remaining empty corner is 0 V?

I thought that this is what i should do. But I cant get the correct answer...

the total potential is
V = k[7.74ƒÊC/r + 7.74ƒÊC/�ã2 *r + q/r]
0 = k/a[(7.74ƒÊC+q)�ã2 + 7.74ƒÊC]
or (7.74ƒÊC+q)�ã2 = - 7.74ƒÊC
q = -7.74ƒÊC[1+1/�ã2]

So that gives me q=-13.23. I am told that that is the wrong answer. Please help!

This looks like a question posted earlier, but in less legible form. I believe BobPursley agreed with your answer. I also got the same answer.

To find the charge magnitude and algebraic sign required at the empty corner of the square, you need to consider the principle of superposition. This principle states that the total potential at a point due to a collection of charges is the algebraic sum of the individual potentials at that point due to each charge separately.

Let's assume that each side of the square has a length of "a". The distance between the corner charge and the empty corner, which we are interested in, is "r". Since the corner charges are identical and fixed to adjacent corners, the distance between them is also "r".

Using Coulomb's Law, the electric potential at the empty corner due to each of the corner charges is given by:

V1 = kq/r (1)

V2 = kq/(√2a) (2)

Note that "k" is the electrostatic constant.

Now, for the total potential at the empty corner to be zero, we have to set the sum of these two potentials equal to zero:

V1 + V2 = 0

kq/r + kq/(√2a) = 0

q/r + q/(√2a) = 0

To simplify the equation, let's divide through by "q":

1/r + 1/(√2a) = 0

Now, let's solve this equation for "r":

1/r = -1/(√2a)

Taking the reciprocal of both sides:

r = -√2a

Since the distance cannot be negative, we discard the negative sign:

r = √2a

Now, substitute this value of "r" into equation (1):

q/√2a + q/(√2a) = 0

2q/(√2a) = 0

2q = 0

q = 0

Therefore, the magnitude and algebraic sign of the charge that should be fixed to one of the empty corners is 0 C. This means that no charge needs to be fixed to the empty corner in order to achieve a total potential of 0 V at the remaining empty corner.