Question

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

Answer 1

(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

Step-by-step 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;

`F = (k \cdot q_1 \cdot q_2)/(d^2)`

Therefore, we have;

`0.0988 \ N = -(8.99 * 10^9 N\cdot m^2 \cdot C^(-2)\cdot q_1 \cdot q_2)/((0.445 \ m)^2)`

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;

`F_2 = (k \cdot \left ((q_1 + q_2)/(2) \right) ^2)/(d^2)`

Therefore, we have;

`q_1 + q_2 = \sqrt{(4 \cdot F_2 \cdot d^2)/(k) }`

`q_1 + q_2 = \sqrt{(4 * 0.0276 \ N *(0.445 \ m)^2)/(8.99 * 10^9 N\cdot m^2 \cdot C^(-2)) } = 1.59446902743 * 10^(-6) \ C`

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