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-Dominant- [34]
3 years ago
14

A point charge of +3 C is located at the origin of a coordinate system and a second point charge of -6 C is at x = 1.0 m. At w

hat point on the x-axis is the electrical potential zero?
Physics
1 answer:
Butoxors [25]3 years ago
8 0

Answer:

The point at which the electrical potential is zero is x = +0.33 m.

Explanation:

By definition the electrical potential is:

V_{E} = \frac{K*q}{r}

Where:

K: is Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

r: is the distance

The point at which the electrical potential is zero can be calculated as follows:

V_{1} + V_{2} = 0

K(\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}}) = 0    (1)

q₁ is the first charge = +3 mC

r₁ is the distance from the point to the first charge  

q₂ is the first charge = -6 mC

r₂ is the distance from the point to the second charge    

By replacing r₁ = 1 - r₂ into equation (1) we have:

K(\frac{q_{1}}{1 - r_{2}} + \frac{q_{2}}{r_{2}}) = 0   (2)

By solving equation (2) for r₂:

r_{2} = \frac{q_{1}}{q_{1} - q_{2}} = \frac{3 mC}{3 mC - (-6 mC)} = +0.33 m

                 

Therefore, the point at which the electrical potential is zero is x = +0.33 m.

I hope it helps you!  

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Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
3 years ago
A 65-kg person stands on a scale in a moving elevator while holding a 5.0 kg mass suspended from a massless spring with spring c
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Answer:

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Explanation:

By Hooke's law, F = ke

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From the question,

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F = ma

a = \dfrac{F}{m}

a = \dfrac{64.8 \text{ N}}{5\text{ kg}} = 12.96\text{ m/s}^2

This acceleration is more than the acceleration due to gravity, g = 9.8 m/s². Hence the elevator must be moving up with an acceleration of

12.96 - 9.8 m/s² = 3.16 m/s²

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