Electric Potential and charge problem

In summary, in order for the potential to be zero at each of the two empty corners of a square with a fixed charge of +9q at one corner and a fixed charge of -8q at the opposite corner, a charge of -q√2/2 should be fixed at the center of the square. This can be expressed as q1 = -q√2/2, where d is the distance between the two positive charges and the distance between q1 and the two empty corners is .5d√2.
  • #1
Biosyn
115
0

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 = [itex]\frac{kq}{r}[/itex]

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

[itex]\frac{k(+9q)}{d}[/itex] + [itex]\frac{k(-8q)}{d}[/itex] + [itex]\frac{k(q1)}{d√2}[/itex] = 0

[itex]\frac{k(+9q-8q)}{d}[/itex] + [itex]\frac{k(q1)}{d√2}[/itex] = 0

[itex]\frac{k(+9q-8q)}{d}[/itex] = -[itex]\frac{k(q1)}{d√2}[/itex]

q1 = -q√2 The answer in the back of the book is [itex]-q/√2[/itex]
 
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  • #2
Biosyn said:

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 = [itex]\frac{kq}{r}[/itex]

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

[itex]\frac{k(+9q)}{d}[/itex] + [itex]\frac{k(-8q)}{d}[/itex] + [itex]\frac{k(q1)}{d√2}[/itex] = 0

[itex]\frac{k(+9q-8q)}{d}[/itex] + [itex]\frac{k(q1)}{d√2}[/itex] = 0

[itex]\frac{k(+9q-8q)}{d}[/itex] = -[itex]\frac{k(q1)}{d√2}[/itex]

q1 = -q√2

The answer in the back of the book is [itex]-q/√2[/itex]
"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.)
 
  • #3
SammyS said:
"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..
 
  • #4
Biosyn said:
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.
 
  • #5
SammyS said:
Yes, that's the distance.

Re-do your previous attempt.

Would the equation be this? :/

[itex]\frac{k(+9q)}{d}[/itex] + [itex]\frac{k(-8q)}{d}[/itex] + [itex]\frac{k(q1)}{.5d√2}[/itex] = 0I get q1 = [itex]\frac{-q√2}{2}[/itex]
 
Last edited:

Related to Electric Potential and charge problem

1. What is electric potential?

Electric potential is the amount of electric potential energy that a charged particle has per unit charge at a given point in an electric field. It is a scalar quantity and is measured in volts (V).

2. How is electric potential different from electric field?

Electric potential is a measure of the potential energy that a charged particle has, while electric field is the force that a charged particle experiences at a given point in an electric field. Electric potential is a scalar quantity, while electric field is a vector quantity.

3. What is the relationship between electric potential and charge?

The electric potential at a point is directly proportional to the amount of charge present at that point. This means that the greater the charge, the higher the electric potential at that point.

4. How is electric potential calculated?

Electric potential can be calculated by dividing the electric potential energy of a charged particle by its charge. The formula for electric potential is V = U/Q, where V is the electric potential, U is the electric potential energy, and Q is the charge.

5. What is the difference between electric potential and electric potential energy?

Electric potential is the potential energy per unit charge at a given point in an electric field, while electric potential energy is the energy that a charged particle has due to its position in an electric field. Electric potential is a measure of the potential energy, while electric potential energy is the actual energy.

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