Respuesta :
Answer:
[tex]\Delta V=316167V[/tex]
Explanation:
The difference of electric potential between two points is given by the formula [tex]\Delta V=Ed[/tex], where d is the distance between them and E the electric field in that region, assuming it's constant.
The electric field formula is [tex]E=\frac{F}{q}[/tex], where F is the force experimented by a charge q placed in it.
Putting this together we have [tex]\Delta V=\frac{Fd}{q}[/tex], so we need to obtain the electric force the charged ball is experimenting.
On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight W, giving a net force [tex]N=W-F[/tex]. On the first drop only W acts, while on the second drop is N that acts.
Using the equation for accelerated motion (departing from rest) [tex]d=\frac{at^2}{2}[/tex], so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:
[tex]a_1=\frac{2d}{t_1^2}[/tex]
[tex]a_2=\frac{2d}{t_2^2}[/tex]
These relate with the forces by Newton's 2nd Law:
[tex]W=ma_1[/tex]
[tex]N=ma_2[/tex]
Putting all together:
[tex]N=W-F=ma_1-F=ma_2[/tex]
Which means:
[tex]F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})[/tex]
And finally we substitute:
[tex]\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})[/tex]
Which for our values means:
[tex]\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V[/tex]
The electric potential at a height of 1.00m above the ground on this planet is 316,145.4545 V.
Given to us
Mass of the ball, m = 0.210 kg
Height, h = 1 m
Time is taken to reach the ground when it was not charged, t₁ = 0.350 s
Time is taken to reach the ground when it was charged, t₂ = 0.650 s
The net charge of the ball = 7.70 μC = 7.7 x 10 ⁻⁶ C
What is the acceleration of the ball on the planet?
According to the second equation of motion,
[tex]S=ut+\dfrac{1}{2}at^2[/tex]
since, the ball is dropping from the position of rest,
[tex]S=\dfrac{1}{2}at^2\\\\[/tex]
Now,
[tex]a_1 = \dfrac{2}{0.350^2}= 16.326 \rm\ m/s^2[/tex]
[tex]a_2 = \dfrac{2}{0.650^2}= 4.734\rm\ m/s^2[/tex]
What is the force at the ball?
According to Newton's second law, Force can be written as
W = ma₁ = 3.42846 N
N = ma₂ = 0.99414 N
F = W- N = 2.43432 N
What is the difference in electric potential?
We know that the difference of electric potential between any two points is given by the formula ΔV =Ed.
We also know that the electric field is given by the formula,
[tex]\overrightarrow E = \dfrac{F}{q}[/tex]
Therefore,
[tex]\triangle V = \overrightarrow Ed = \dfrac{Fd}{q}[/tex]
[tex]\triangle V = \dfrac{2.43432 \times 1}{7.7 \times 10^{-6}} =316,145.4545\rm\ V[/tex]
Hence, the electric potential at a height of 1.00m above the ground on this planet is 316,145.4545 V.
Learn more about Electric potential:
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