Question

A fixed particle with charge –8.8 nC and a second particle with charge –4.3 nC, respectively, are initially separated by a distance of 0.11 m. They are released and the second particle moves 0.030 m. A) What is the change in electric potential energy? B) Did the second particle move toward or away from the source charge?

Answer 1

Answer:

(a). The change in electric potential energy is 8.3 μJ.

(b).  The second particle moves away from the source charge.

Explanation:

Given that,

Charge of first particle = -8.8 nC

Charge of second particle = -4.3 nC

Distance = 0.11 m

They are released and the second particle moves 0.030 m,

(a). We need to calculate the change in electric potential energy

Using formula of potential energy

$U=\dfrac{Kq_{1}q_{2}}{d}$

Change in potential energy

$\Delta U=\dfrac{kq_{1}q_{2}}{d_{2}}-\dfrac{Kq_{1}q_{2}}{d_{1}}$

Put the value into the formula

$U=\dfrac{9\times10^{9}\times(8.8\times10^{-9}\times4.3\times10^{-9})}{0.030}-\dfrac{9\times10^{9}\times(8.8\times10^{-9}\times4.3\times10^{-9})}{0.11}$

$\Delta U=0.000008256\ J$

$\Delta U=8.3\ \mu J$

(B). We need to find the second particle move toward or away from the source charge

We know that,

Both charges are same, so the second particle will be repul from the source charge.

So, The second particle moves away from the source charge.

Hence, (a). The change in electric potential energy is 8.3 μJ.

(b).  The second particle moves away from the source charge.