• Shatzkinator
In summary, two small, positively charged spheres with a combined charge of 14.0 × 10-5 C are repelled from each other by an electrostatic force of 0.9 N when they are 1.7 m apart. Using the equations F = kq1q2/d^2 and q2 = 14 x 10^-5 - q1, a quadratic equation can be solved to find the charge on the smaller sphere. However, to get an accurate answer, it is important to not round off any numbers until the final answer, which is calculated to be 2.10 x 10^-6 C.

## Homework Statement

Two small, positively charged spheres have a combined charge of 14.0 × 10-5 C. If each sphere is repelled from the other by an electrostatic force of 0.9 N when the spheres are 1.7 m apart, what is the charge (in Coulombs) on the sphere with the smaller charge?

## Homework Equations

(-b +/- sqt (b^2 - 4ac))/2a
F = kq1q2/d^2
q2 = 14 x 10^-5 - q1 --> substitute for q2 in above equation

## The Attempt at a Solution

I solved the quadratic equation to get 2.22 x 10^-6 and 1.38 x 10^-4 .. neither answer is correct. Help?

Hi Shatzkinator!

Show us your full calculations, and then we can see what went wrong, and we'll know how to help!

0.9 = (8.99 x 10^9)q1 (14 x 10^-5 - q1)/ 1.7^2
2.6 = 1.26 x 10^6 q1 - 8.99 x10^9 q1^2
-8.99 x10^9 q1^2 + 1.26 x 10^6 q1 - 2.6 = 0

-1.26 x 10^6 + sqt ((1.26 x 10^6)^2 - 4(- 8.99^9)(-2.6)) / 2(-8.99^9) = 2.22 x 10^6

the subtraction root from same equation = 1.38 x10^-4

Hi Shatzkinator!

(try using the X2 tag just above the Reply box )
Shatzkinator said:
… -8.99 x10^9 q1^2 + 1.26 x 10^6 q1 - 2.6 = 0

-1.26 x 10^6 + sqt ((1.26 x 10^6)^2 - 4(- 8.99^9)(-2.6)) / 2(-8.99^9) = 2.22 x 10^6

the subtraction root from same equation = 1.38 x10^-4

hmm … looks ok, but there must be a mistake somewhere …

the two solutions have to add up to 14 x 10-5 C

i checked over calculations and everything... really stumped

thkz for the 103 tip

I make the smaller charge nearer 1.8 x 10-6

it's very much smaller than the other one …

maybe you're rounding off too much?

1.8 x 10-6 is incorrect as well :(

my answer is 2.22 x 10^-6 not 6
both are incorrect anyways

Shatzkinator said:
1.8 x 10-6 is incorrect as well :(

my answer is 2.22 x 10^-6 not 6
both are incorrect anyways

I get $2.10\times10^{-6}\text{C}$. Try not rounding any of your numbers off until the very end of your calculations.

## 1. What are charged spheres and why do we use a quadratic equation to study them?

Charged spheres are objects with a net electric charge. We use a quadratic equation to study them because the electric field and potential of a charged sphere follows a quadratic relationship with distance.

## 2. How do we solve for the electric potential of a charged sphere using a quadratic equation?

To solve for the electric potential, we use the equation V = kq/r, where V is the electric potential, k is the Coulomb's constant, q is the charge of the sphere, and r is the distance from the center of the sphere. This equation can be rearranged to form a quadratic equation, which can be solved using the quadratic formula.

## 3. Can a charged sphere have a negative electric potential?

Yes, a charged sphere can have a negative electric potential if it has a negative charge. The electric potential is a scalar quantity, meaning it can be positive or negative depending on the charge of the sphere.

## 4. How does the electric field of a charged sphere vary with distance?

The electric field of a charged sphere varies inversely with the square of the distance from the center of the sphere. This means that as the distance increases, the electric field decreases. It follows a similar relationship to that of the electric potential.

## 5. Are there any real-world applications of studying charged spheres using a quadratic equation?

Yes, there are many real-world applications of studying charged spheres using a quadratic equation. Some examples include understanding the behavior of charged particles in an electric field, designing electric motors and generators, and calculating the potential and electric field of charged conductors.