To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:
= Coulomb's constant
q = Charge
r = Radius
At the same time
The values of variables are the same, then if we replace in a single equation we have this expression,
If we replace the values, we have finally that the charge is,
Therefore the potential energy of the system is 
Answer:
1 / i + 1 / o = 1 / f thin lens equations
i = o f / (o - f) rearranging
Lens 1: object = 30 cm f = 15.2 cm
i1 = 30 * 15.2 / (30 - 15.2) = 30.8 cm
o2 = 40.2 - 30/8 = 9.4 cm distance of image 1 from lens 2
i2 = 9.4 * 15.2 / (9.4 - 15.2) = - 24.6 cm
The final image is 24.6 cm to the left of lens 2
The first image is inverted
The second image is erect (as seen from the first image)
So the final image is inverted
M = m1 * m2 = (-30.8 / 30) * (24.6 / 9.4) = -2.69
The angle of reflection is the angle that the reflected ray makes with a line drawn perpendicular to the reflecting surface.
