Potential energy is sometimes called?
2 answers:
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Answer:
The charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C
Explanation:
Here is the complete question
Two identical tiny balls have charge q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 X 10-4 N. after the balls are touched together and then represented once again to 20cm, now the repulsive force is found to be 1.40 X 10-4 N. find the charges q1 and q2.
Solution
The force F = 1.35 × 10⁻⁴ N when the charges are separated a distance of r = 20 cm = 0.2 m is given by
F = kq₁q₂/r₁²
q₁q₂ = Fr₁²/k
q₁q₂ = 1.35 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.054/9 × 10⁻¹³ C² = 0.006 × 10⁻¹³ C² = 6 × 10⁻¹⁶ C²
q₁q₂ = 6 × 10⁻¹⁶ C² (1)
When the charges are brought together, the charge is now q = (q₁ + q₂)/2
The new repulsive force F = 1.406 × 10⁻⁴ N at a distance of r₂ = 20 cm = 0.2 m is then
F₂ = kq²/r₂²
q² = F₂r₂²/k = 1.406 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.00625 × 10⁻¹³ C² = 6.25 × 10⁻¹⁶ C²
q² = 6.25 × 10⁻¹⁶ C²
q = √(6.25 × 10⁻¹⁶) C
q = 2.5 × 10⁻⁸ C
(q₁ + q₂)/2 = 2.5 × 10⁻⁸ C
(q₁ + q₂) = 2 × 2.5 × 10⁻⁸ C
q₁ + q₂ = 5 × 10⁻⁸ C (2)
q₁ = 5 × 10⁻⁸ C - q₂ (3)
Substituting equation (3) into (1), we have
(5 × 10⁻⁸ C - q₂)q₂ = 6 × 10⁻¹⁶ C²
Expanding the bracket, we have
(5 × 10⁻⁸ C)q₂ - q₂² = 6 × 10⁻¹⁶ C²
So, q₂² - (5 × 10⁻⁸ C)q₂ + 6 × 10⁻¹⁶ C² = 0
Using the quadratic formula to find q₂
q₁ = 5 × 10⁻⁸ C - q₂
q₁ = 5 × 10⁻⁸ C - 3 × 10⁻⁸ C or 5 × 10⁻⁸ C - 2 × 10⁻⁸ C
q₁ = 2 × 10⁻⁸ C or 3 × 10⁻⁸ C
So the charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C
The elastic potential energy of a spring is given by
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.
The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
In our problem, initially the spring is uncompressed, so
. Therefore, the work done by the spring when it is compressed until 
is
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.
The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
In our problem, initially the spring is uncompressed, so
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
Answer:
Explanation:
The formula for the force exerted between two charges is
where k is the Coulomb constant.
The charges are identical, so we can write the formula as