Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.0988 N when their center-to-center separation is 44.5 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0276 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other

(a) The negative charge on one of the charges is -8.79630245 × 10⁻⁷C

(b) The positive charge on one of the other charges is 8.79630245 × 10⁻⁷C

Explanation:

The given parameters are;

The force of attraction between the two spheres = 0.0988 N

The distance between their centers = 44.5 cm = 0.445 m

Therefore, we have;

Therefore, we have;

Therefore, we have;

q₁·q₂ = -0.0988 N × (0.445 m)²/(8.99 × 10⁹ N·m²·C⁻²) = -2.17629255 × 10⁻¹² C²

q₁·q₂ = -2.17629255 × 10⁻¹² C²...(1)

When the two charges are connected, we get;

Therefore, we have;

q₁ + q₂ = 1.59446902743 × 10⁻⁶ C...(2)

From, equation (2), we have;

q₁ = 1.59446902743 × 10⁻⁶ C -  q₂

Plugging in the value of q₁ in equation (1) givens;

q₁·q₂ = -2.17629255 × 10⁻¹²

Therefore, we have;

(1.59446902743 × 10⁻⁶ -  q₂) × q₂ = -2.17629255 × 10⁻¹²

Which gives;

-q₂² + 1.59446902743 × 10⁻⁶·q₂+2.17629255 × 10⁻¹² = 0

Solving, with a graphing calculator, we get;

q₂ = 2.4741×10⁻⁶ C, or -8.79630245 × 10⁻⁷C

q₁ = 8.79630245 × 10⁻⁷C or -2.4741×10⁻⁶ C

Therefore, we have;

(a) The negative charge on one of the charges = -8.79630245 × 10⁻⁷C

(b) The positive charge on one of the other charges = 8.79630245 × 10⁻⁷C

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2