Answer:
Option B. 3.0×10¯¹¹ F.
Explanation:
The following data were obtained from the question:
Potential difference (V) = 100 V.
Charge (Q) = 3.0×10¯⁹ C.
Capacitance (C) =..?
The capacitance, C of a capacitor is simply defined as the ratio of charge, Q on either plates to the potential difference, V between them. Mathematically, it is expressed as:
Capacitance (C) = Charge (Q) / Potential difference (V)
C = Q/V
With the above formula, we can obtain the capacitance of the parallel plate capacitor as follow:
Potential difference (V) = 100 V.
Charge (Q) = 3.0×10¯⁹ C.
Capacitance (C) =..?
C = Q/V
C = 3.0×10¯⁹ / 100
C = 3.0×10¯¹¹ F.
Therefore, the capacitance of the parallel plate capacitor is 3.0×10¯¹¹ F.
Given :
An electron moving in the positive x direction experiences a magnetic force in the positive z direction.
To Find :
The direction of the magnetic field.
Solution :
We know, force is given by :
Here, q = -e.
Now, for above condition to satisfy :
So, 
Therefore, direction of magnetic field is negative y direction.
Hence, this is the required solution.
The answer to what experimental result required considering the particle nature of light is A. The ultraviolet catastrophe of blackbody radiation.
