What Is the Acceleration of a Charged Sphere Fired Toward a Fixed Charge?

[miniake]

[SOLVED] Question about Electric field.

Homework Statement

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

Homework Equations

a = \frac{k*q_{1}*q_{2}}{m*r^{2}}

The Attempt at a Solution

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

a = 67.5 * \frac{1}{r^{2}}

By integration,

v = -67.5 * \frac{1}{r} + C

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

38 = -67.5 * \frac{1}{0.03} + C

C = 2288

v = -67.5 * \frac{1}{r} + 2288

When v = 20m/s,

20 = -67.5 * \frac{1}{r} + 2288

r = 0.029761904

a = 67.5 * \frac{1}{0.029761904^{2}}

a ≈ 76200 m/s^{2}

Since both charges are +ve,

a = -76200 m/s^{2}

However, the solution is not correct.

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

 

[gneill]

Homework Statement

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

Homework Equations

a = \frac{k*q_{1}*q_{2}}{m*r^{2}}

The Attempt at a Solution

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

a = 67.5 * \frac{1}{r^{2}}

By integration,

v = -67.5 * \frac{1}{r} + C

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

38 = -67.5 * \frac{1}{0.03} + C

C = 2288

v = -67.5 * \frac{1}{r} + 2288

When v = 20m/s,

20 = -67.5 * \frac{1}{r} + 2288

r = 0.029761904

a = 67.5 * \frac{1}{0.029761904^{2}}

a ≈ 76200 m/s^{2}

Since both charges are +ve,

a = -76200 m/s^{2}

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 ?

 

[Tanya Sharma]

Hello miniake

What is the correct answer ?

Edit :Conservation of energy is the correct way to approach the problem.

 

[miniake]

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 U_{1}+KE_{1} = U_{2}+KE_{2}.

Thanks again.

Hello miniake

What is the correct answer ?

Using the above approach will work.