Electric Potential and charge problem

[Biosyn]

Homework Statement

A charge of +9q is fixed to one corner of a square, while a charge of -8q is fixed to the opposite corner. Expressed in terms of q, what charge should be fixed to the center of the square, so the potential is zero at each of the two empty corners?

Homework Equations

V = $\frac{kq}{r}$

The Attempt at a Solution

q1 = charge that should be fixed to center

d = distance between the two positive charges
The distance that the charge in the center of the square is from the two other charges is: d√2

$\frac{k(+9q)}{d}$ + $\frac{k(-8q)}{d}$ + $\frac{k(q1)}{d√2}$ = 0

$\frac{k(+9q-8q)}{d}$ + $\frac{k(q1)}{d√2}$ = 0

$\frac{k(+9q-8q)}{d}$ = -$\frac{k(q1)}{d√2}$

q1 = -q√2 The answer in the back of the book is $-q/√2$

 

[SammyS]

Homework Statement

A charge of +9q is fixed to one corner of a square, while a charge of -8q is fixed to the opposite corner. Expressed in terms of q, what charge should be fixed to the center of the square, so the potential is zero at each of the two empty corners?

Homework Equations

V = $\frac{kq}{r}$

The Attempt at a Solution

q1 = charge that should be fixed to center

d = distance between the two positive charges
The distance that the charge in the center of the square is from the two other charges is: d√2

$\frac{k(+9q)}{d}$ + $\frac{k(-8q)}{d}$ + $\frac{k(q1)}{d√2}$ = 0

$\frac{k(+9q-8q)}{d}$ + $\frac{k(q1)}{d√2}$ = 0

$\frac{k(+9q-8q)}{d}$ = -$\frac{k(q1)}{d√2}$

q1 = -q√2

The answer in the back of the book is $-q/√2$

"The distance that the charge in the center of the square is from the two other charges is: d√2 ."

d√2 > d . How can that be?

What you really want is the distance from q₁ to the two empty corners of the square. (Of course that is equal to the distance from q₁ to each of the two other charges.)

 

[Biosyn]

"The distance that the charge in the center of the square is from the two other charges is: d√2 ."

d√2 > d . How can that be?

What you really want is the distance from q₁ to the two empty corners of the square. (Of course that is equal to the distance from q₁ to each of the two other charges.)

Would that distance be .5d√2 ?

I'm not really sure how to set up an equation for this problem..

 

[SammyS]

Would that distance be .5d√2 ?

I'm not really sure how to set up an equation for this problem..

Yes, that's the distance.

Re-do your previous attempt.

 

[Biosyn]

Yes, that's the distance.

Re-do your previous attempt.

Would the equation be this? :/

$\frac{k(+9q)}{d}$ + $\frac{k(-8q)}{d}$ + $\frac{k(q1)}{.5d√2}$ = 0I get q₁ = $\frac{-q√2}{2}$