Question about Electric field.

[SOLVED] Question about Electric field.

1. The problem statement, all variables and given/known data
A point charge [itex]q_{1}=15.00\mu C[/itex] is held fixed in space. From a horizontal distance of [itex]3.00cm[/itex] , a small sphere with mass [itex]4.00*10^{-3}kg[/itex] and charge [itex]q_{2}=+2.00\mu[/itex]C is fired toward the fixed charge with an initial speed of [itex]38.0m/s[/itex] . Gravity can be neglected.
What is the acceleration of the sphere at the instant when its speed is [itex]20.0m/s[/itex] ?
Express your answer with the appropriate units.

2. Relevant equations

[itex]a = \frac{k*q_{1}*q_{2}}{m*r^{2}}[/itex]


3. The attempt at a solution

[itex]a = \frac{9*10^{9}*15*10^{-6}*2*10^{-6}}{4*10^{-3}*r^{2}}[/itex]

[itex]a = 67.5 * \frac{1}{r^{2}}[/itex]

By integration,

[itex]v = -67.5 * \frac{1}{r} + C[/itex]

Initial conditions: [itex]v = 38 m/s, r = 0.03m [/itex]

[itex]38 = -67.5 * \frac{1}{0.03} + C[/itex]

[itex]C = 2288[/itex]

[itex]v = -67.5 * \frac{1}{r} + 2288[/itex]

When [itex]v = 20m/s[/itex],

[itex]20 = -67.5 * \frac{1}{r} + 2288[/itex]

[itex]r = 0.029761904[/itex]

[itex]a = 67.5 * \frac{1}{0.029761904^{2}}[/itex]

[itex]a ≈ 76200 m/s^{2}[/itex]

Since both charges are +ve,

[itex]a = -76200 m/s^{2}[/itex]

However, the solution is not correct.

May anyone pointing out the errors? Thank you very much.
 
Last edited:

gneill

Mentor
20,400
2,568
1. The problem statement, all variables and given/known data
A point charge [itex]q_{1}=15.00\mu C[/itex] is held fixed in space. From a horizontal distance of [itex]3.00cm[/itex] , a small sphere with mass [itex]4.00*10^{-3}kg[/itex] and charge [itex]q_{2}=+2.00\mu[/itex]C is fired toward the fixed charge with an initial speed of [itex]38.0m/s[/itex] . Gravity can be neglected.
What is the acceleration of the sphere at the instant when its speed is [itex]20.0m/s[/itex] ?
Express your answer with the appropriate units.

2. Relevant equations

[itex]a = \frac{k*q_{1}*q_{2}}{m*r^{2}}[/itex]


3. The attempt at a solution

[itex]a = \frac{9*10^{9}*15*10^{-6}*2*10^{-6}}{4*10^{-3}*r^{2}}[/itex]

[itex]a = 67.5 * \frac{1}{r^{2}}[/itex]

By integration,

[itex]v = -67.5 * \frac{1}{r} + C[/itex]

Initial conditions: [itex]v = 38 m/s, r = 0.03m [/itex]

[itex]38 = -67.5 * \frac{1}{0.03} + C[/itex]

[itex]C = 2288[/itex]

[itex]v = -67.5 * \frac{1}{r} + 2288[/itex]

When [itex]v = 20m/s[/itex],

[itex]20 = -67.5 * \frac{1}{r} + 2288[/itex]

[itex]r = 0.029761904[/itex]

[itex]a = 67.5 * \frac{1}{0.029761904^{2}}[/itex]

[itex]a ≈ 76200 m/s^{2}[/itex]

Since both charges are +ve,

[itex]a = -76200 m/s^{2}[/itex]

However, the solution is not correct.

May anyone pointing out the errors? Thank you very much.
Hi miniake, Welcome to Physics Forums.

I think the problem lies with your assumption that you can integrate the acceleration with respect to distance to yield velocity. Velocity is the integral of acceleration with respect to time.

Have you considered using a conservation of energy approach to find the separation for v = 20.0 m/s ?
 
1,540
133
Hello miniake

What is the correct answer ?

Edit :Conservation of energy is the correct way to approach the problem.
 
Last edited:
Hi miniake, Welcome to Physics Forums.

I think the problem lies with your assumption that you can integrate the acceleration with respect to distance to yield velocity. Velocity is the integral of acceleration with respect to time.

Have you considered using a conservation of energy approach to find the separation for v = 20.0 m/s ?
Thank you gneill.

It works, using the formula [itex]U_{1}+KE_{1} = U_{2}+KE_{2}[/itex].

Thanks again.

Hello miniake

What is the correct answer ?
Using the above approach will work.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top