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 $q_{2}=+2.00\mu$C 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 $q_{2}=+2.00\mu$C 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.

 

[HalfLostAndSearching]

Your approach to solving this problem is correct, but there are a few errors in your calculations. First, the value of k should be 8.99*10^9, not 9*10^9. Second, the charge of q1 should be 1.5*10^-5C, not 15*10^-6C. Third, in the equation for acceleration, the value of r should be squared, not just the unit. Finally, in the integration step, the constant should be -67.5, not 2288. With these corrections, you should be able to get the correct answer. Keep up the good work!