Answer:
c
Explanation:
I think but if I'm wrong I'm sorry
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.
Answer:
hello the diagram related to this question is missing attached below is the missing diagram
Answer :
The magnitude of the electric field = 4KQ / L^2
direction = 45° east to south
Explanation:
The magnitude of the electric field = 4KQ / L^2
direction = 45° east to south
Answer:
![\Delta E=2.87\times 10^6\ J](https://tex.z-dn.net/?f=%5CDelta%20E%3D2.87%5Ctimes%2010%5E6%5C%20J)
Explanation:
It is given that,
Depth of Death valley is 85 m below sea level, ![h_i=-85\ m](https://tex.z-dn.net/?f=h_i%3D-85%5C%20m)
The summit of nearby Mt. Whitney has an elevation of 4420 m, ![h_f=4420\ m](https://tex.z-dn.net/?f=h_f%3D4420%5C%20m)
Mass of the hiker, m = 65 kg
We need to find the change in potential energy. It is given by :
![\Delta E=mg(h_f-h_i)](https://tex.z-dn.net/?f=%5CDelta%20E%3Dmg%28h_f-h_i%29)
![\Delta E=65\times 9.8(4420-(-85))](https://tex.z-dn.net/?f=%5CDelta%20E%3D65%5Ctimes%209.8%284420-%28-85%29%29)
![\Delta E=2869685\ J](https://tex.z-dn.net/?f=%5CDelta%20E%3D2869685%5C%20J)
or
![\Delta E=2.87\times 10^6\ J](https://tex.z-dn.net/?f=%5CDelta%20E%3D2.87%5Ctimes%2010%5E6%5C%20J)
So, the change in potential energy of the hiker is
. Hence, this is the required solution.
The last one, handmade gifts require more of the givers time!