help, please ! physics ?
2020-08-19 1:25 am
the electrostatic force between two point charges is 2.8 x 10^-5 N. If the distance between them doubles, the charge of one of the points quadruples and the charge of the other point is cut by a factor of 3. What will the new force between them be?
回答 (2)
2020-08-19 1:37 am
Original electrostatic force:
kₑ q₁ q₂ / r² = 2.8 × 10⁻⁵ N
When the system becomes 2r, 4q₁ and q₂/3, new force
= kₑ (4q₁) (q₂/3) / (2r)²
= (1/3) (kₑ q₁ q₂ / r²)
= (1/3) × (2.8 × 10⁻⁵ N)
= 9.3 × 10⁻⁶ N
kₑ q₁ q₂ / r² = 2.8 × 10⁻⁵ N
When the system becomes 2r, 4q₁ and q₂/3, new force
= kₑ (4q₁) (q₂/3) / (2r)²
= (1/3) (kₑ q₁ q₂ / r²)
= (1/3) × (2.8 × 10⁻⁵ N)
= 9.3 × 10⁻⁶ N
2020-08-19 1:37 am
doubling the distance cuts the force by a factor of 4
Quadrupling a charge multiples the force by 4
cutting a charge by 1/3 reduces the force by that.
overall:
(1/4)(4)(1/3) = 1/3
(1/3)(2.8 x 10^-5 N) = _____ N
Quadrupling a charge multiples the force by 4
cutting a charge by 1/3 reduces the force by that.
overall:
(1/4)(4)(1/3) = 1/3
(1/3)(2.8 x 10^-5 N) = _____ N
收錄日期: 2021-04-24 07:58:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200818172501AAu897G