Using the formula F=ma
500N=50kg (a)
a= 10 m/s^2
Answer:
y = 10.2 m
Explanation:
It is given that,
Charge, ![q_1=-3\ nC](https://tex.z-dn.net/?f=q_1%3D-3%5C%20nC)
It is placed at a distance of 9 cm at x axis
Charge, ![q_2=+4\ nC](https://tex.z-dn.net/?f=q_2%3D%2B4%5C%20nC)
It is placed at a distance of 16 cm at x axis
We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,
![\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0](https://tex.z-dn.net/?f=%5Cdfrac%7Bkq_1%7D%7Br_1%7D%2B%5Cdfrac%7Bkq_2%7D%7Br_2%7D%3D0)
Here,
![r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}](https://tex.z-dn.net/?f=r_1%3D%5Csqrt%7By%5E2%2B9%5E2%7D%20%5C%5C%5C%5Cr_2%3D%5Csqrt%7By%5E2%2B15%5E2%7D)
So,
![\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}](https://tex.z-dn.net/?f=%5Cdfrac%7Bkq_1%7D%7Br_1%7D%3D-%5Cdfrac%7Bkq_2%7D%7Br_2%7D%5C%5C%5C%5C%5Cdfrac%7Bq_1%7D%7Br_1%7D%3D-%5Cdfrac%7Bq_2%7D%7Br_2%7D%5C%5C%5C%5C%5Cdfrac%7B-3%5C%20nC%7D%7B%5Csqrt%7By%5E2%2B81%7D%20%7D%3D-%5Cdfrac%7B4%5C%20nC%7D%7B%5Csqrt%7By%5E2%2B225%7D%20%7D%5C%5C%5C%5C3%5Ctimes%20%5Csqrt%7By%5E2%2B225%7D%3D4%5Ctimes%20%5Csqrt%7By%5E2%2B81%7D)
Squaring both sides,
![3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m](https://tex.z-dn.net/?f=3%5Ctimes%20%5Csqrt%7By%5E2%2B225%7D%3D4%5Ctimes%20%5Csqrt%7By%5E2%2B81%7D%5C%5C%5C%5C9%28y%5E2%2B225%29%3D16%5Ctimes%20%28y%5E2%2B81%29%5C%5C%5C%5C9y%5E2%2B2025%3D16y%5E2-%2B1296%5C%5C%5C%5C2025-1296%3D7y%5E2%5C%5C%5C%5C7y%5E2%3D729%5C%5C%5C%5Cy%3D10.2%5C%20m)
So, at a distance of 10.2 m on the y axis the electric potential equals 0.
Thermal conductions
K= QL/ART
Aluminium T₁ = 10 + 273.15
T₂ = 283.15k
205 = 2.0 × 0.30/4× 10⁻⁴ × (T₂ - 283.15)
Copper
385 = Q × 0.70/4×10⁻⁴ ×(433.15 - T₂)
Where T₃ = 160 + 273.15
T₃ = 433.15K
From 2 to 3
205/385 = 0.30/0.70 × 433.15 - T₂/T₂ - 283.15
= 0.53T₂ -150.06 = 181.92 - 0.42 T₂
→ 0.95T₂ = 331.98 ⇒ T₂ = ₂349.45k
T₂ = 76.3°c
=77°c.
It is required an infinite work. The additional electron will never reach the origin.
In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
![F=k_e \frac{q_e q_e}{d^2}](https://tex.z-dn.net/?f=F%3Dk_e%20%20%5Cfrac%7Bq_e%20q_e%7D%7Bd%5E2%7D%20)
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.