Question

IP A charge of 18.0 μCμC is held fixed at the origin.

Part A:

If a -7.00 μCμC charge with a mass of 3.40 gg is released from rest at the position (0.925 mm, 1.17 mm), what is its speed when it is halfway to the origin?

Part B:

Suppose the -7.00 μCμC charge is released from rest at the point xx = 1212(0.925mm) and yy = 1212(1.17mm). When it is halfway to the origin, is its speed greater than, less than, or equal to the speed found in part A?

Suppose the -7.00  charge is released from rest at the point  = (0.925) and  = (1.17). When it is halfway to the origin, is its speed greater than, less than, or equal to the speed found in part A?

Greater than the speed found in part A

Less than the speed found part in A

Equal to the speed found in part A

Part D

Find the speed of the charge for the situation described in part B.

Answer 1

Given that

Charge located at the origin q1 = 18 C = 18*10^-6 C

Charge located at another position q2  = -7C = -7*10^-6 C

Mass of q2 = 3.40 g = 0.0034 kg

Location of q2 charge x (0.925, 1.17)

Part(A)

We know that

when charge q2 is released from its position x the kinetic energy of q2 will be equal to potential energy of q1 (energy conservation law)

PE of q1 = Kq1q2 / r^2

KE of q2 = 1/2 mv^2

1/2 mv^2 = Kq1q2 / r^2 -------- ------------(1)

Now here r can be calculated as

r = sqrt(0.925^2+1.17^2)

r = 1.49 m

since we have to calculate speed of q2 when it is half way to the origin

R = r /2 = 1.49/2 = 0.745 m

Now putting this value of R in equation (1)

1/2 mv2 = Kq1q2 / (0.745)^2

0.5*0.0034(v)^2 = (9*10^9)(18*10^-6)(-7*10^-6) / (0.745)^2

v = 34.66 m/s

Part(B)

Now the same charge q2 is released from the point x' =12(0.925), y= 12(1.17) (if i understood it correctly)

x' = (11.04 ,14.04)

So

1/2 mv'^2 = Kq1q2 / r^2

we can calculate r as below

R = r/2 = Sqrt(11.04^2 +14.04^2) /2 = 17.86/ 2 = 8.93 m

0.5*0.0034*v'^2 = (9*10^9)(7*10^-6)(18*10^-6)/ 8.93^2

v' = 2.89 m/s

So speed will be less than part A

Part(C)

There are all same values x(0.925, 1.17)  are written in question

Part(D)

already calculated v' = 2.89 m/s

Good Luck !! Do let me know if you have any query