Answer:
A. the left half becomes neutral while the right half remains negatively charged
Explanation:
This is because wherever light strikes the photoconductor, it transforms from an insulator into a conductor. The charge will then migrate through it and leaves its surface. By exposing the left half of the photoconductor to light, you allow its local charge to leave and it becomes neutral.
For these question, it has two separate equations: 2f(a) and f(2a) .
For f(2a) equations its x=2a, so you must substitute 2a into the f(x) equation
For 2f(a), it means the two time of f(a) equation with x=a, so you substitute a inti f(x) equation first, then you multiply it by 2.
Voltage=Energy/Charge, V=E/Q
V in volts V, E in joules J, Q in coulombs C.
or
Voltage= Current×Resistance, V=IR
V in volts V, I in amperes A and R in ohms Ω
