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miniake

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**[SOLVED] Question about Electric field.**

## Homework Statement

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.

## Homework Equations

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

## 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.

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