Physics, 26.03.2020 22:55 HAPPYVIBES

If the third charge (–| q 3 |) is placed at point P, but not held fixed, it will experience a force and accelerate away from the other two charges. If q3 = q = –1.0 nC, the mass of the third charge m3 is 5.0 x 10-12 kg, and s = 4.0 cm, what will the speed of charged particle #3 [in m/s] be after it has moved a very large distance away?

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Answer from: evanwall91

Complete Question

The complete question is shown on the first uploaded image

The potential of this system is $U=6.75*10^{-7}J$

The electric potential at point p is V_p= -900V

The work required is $W= 9*10^{-7}J$

The speed of the charge is v=600m/s

Explanation:

A sketch to explain the question is shown on the second uploaded image

Generally the potential energy for a system of two charges is mathematically represented as

$U = \frac{kq_1 q_2}{d}$

where k is the electrostatic constant with a value of k = 9*10^9 N m^2 /C^2

q is the charge with a value of $q = 1*10^{-9}C$

d is the distance given as d =5m

Now we are given that q_1 = q and q_2 = 3q and

Now substituting values

$U = \frac{9*10^9 *1*10^{-9} * 3*10^{-9}}{5}$

$U=6.75*10^{-7}J$

The electric potential at point P is mathematically obtained with the formula

$V_p = V_{-q} + V_{-3q}$

I.e the potential at q_1 plus the potential at q_2

Now potential at q_1 is mathematically represented as

$V_{-q} = \frac{-kq}{s}$

and the potential at q_2 is mathematically represented as

$V_{-3q} = \frac{-3kq}{s}$

Now substituting into formula for potential at P

$V_p = \frac{-kq}{s} + \frac{-3kq}{s} = -\frac{4kq}{s}$

$= \frac{4*9*10^9 *1*10^{-9}}{4*10^{-2}}$

V_p= -900V

The Workdone to bring the third negative charge is mathematically evaluated as

$W =\Delta U = \frac{kq_1q_3}{s} + \frac{kq_2q_3}{s}$

$= \frac{kq*q}{s} + \frac{kq*3q}{s}$

$= \frac{4kq^2}{s}$

$= \frac{4* 9*10^9 * (1*10^{-9})^2}{4*10^{-2}}$

$W= 9*10^{-7}J$

From the Question are told that the charge q_3 would a force and an acceleration which implies that all its potential energy would be converted to kinetic energy.This can be mathematically represented as

$\Delta U = W = \frac{1}{2} m_{q_3} v^2$

$9*10^{-7} = \frac{1}{2} m_{q_3} v^2$

Where $m_{q_3} = 5.0*10^{-12}kg$

Now making v the subject we have

$v = \sqrt{\frac{9*10^{-12}}{5*10^{-12}*0.5} }$

v=600m/s

If the third charge (–| q 3 |) is placed at point P, but not held fixed, it will experience a force

Answer from: Quest

explanation:

i believe the answer is c and this is only worth 10 points you might wanna change it

20 points! according to classical physics what should happen as a filament in a light bulb gets hot

Answer from: Quest

You would need to take a picture so i can see

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