Two charges q 1 = −3.20 nC and q2 = +8.96 nC are at a distance of 1.94 µm from each other. q 1 is fixed at its location while q 2 is released from rest.

Question

(a) What is the kinetic energy of the charge q 2 when it is 0.340 µm from q 1?
J

(b) The charge q 2 has a mass m 2 = 7.85 µg. What is its speed when it is 0.340 µm from q 1?
m/s

Answer 1

First, we will use conservation of energy to solve part (a). The initial potential energy (PE_initial) is transformed into final potential energy (PE_final) and kinetic energy (KE_final).

PE_initial = KE_final + PE_final

We start by finding the potential energy:

PE = k * q1 * q2 / r

Where k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2)
q1 and q2 are the charges, and r is the distance between the charges.

PE_initial = (8.99 x 10^9) * (-3.20 x 10^-9 C) * (8.96 x 10^-9 C) / (1.94 x 10^-6 m)
PE_initial = -126.82 x 10^-6 J
PE_initial = -126.82 μJ

Now, we find PE_final with the new distance between the charges (0.340 μm):

PE_final = (8.99 x 10^9) * (-3.20 x 10^-9 C) * (8.96 x 10^-9 C) / (0.340 x 10^-6 m)
PE_final = -71.14 x 10^-6 J
PE_final = -71.14 μJ

Now using conservation of energy and solving for KE_final:

KE_final = PE_initial - PE_final
KE_final = -126.82 μJ + 71.14 μJ
KE_final = -55.68 μJ

So the kinetic energy of q2 when it is 0.340 μm from q1 is:

KE_final = -55.68 μJ

Now, we'll find the speed of q2 for part (b). We'll use the formula for kinetic energy:

KE = 0.5 * m * v^2

Where m is the mass of q2 and v is its speed. We will rearrange the formula for v:

v = sqrt (2 * KE / m)

Substitute the values:

v = sqrt (2 * (-55.68 x 10^-6 J) / (7.85 x 10^-6 kg))
v = sqrt (-14.16)
v = -3.765 m/s

But the speed must be positive, so we take the absolute value:

v = 3.765 m/s

Therefore, the speed of q2 when it is 0.340 μm from q1 is 3.765 m/s.