I think the correct answer is the third statement, electron from escaping from the tube containing the triode. The negative charge on the grid repels any electron in the tube. As a result,the flow of current is controlled. If the field is that strong, all current flow will stop resulting to maintaining the electron cloud in the tube.
Answer:
7.9060 m²
8.57 Volts
5.142×10⁻⁶ Joule
1.2×10⁻⁶ Coulomb
Explanation:
C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F
d = Distance between plates = 0.5 mm = 0.5×10⁻³ m
Q = Charge = 1.2 μC = 1.2×10⁻⁶ C
ε₀ = Permittivity = 8.854×10⁻¹² F/m
Capacitance

∴ Area of each plate is 7.9060 m²
Voltage

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.
Energy stored
E=0.5CV²
⇒E = 0.5×0.14×10⁻⁶×8.57²
⇒E = 5.142×10⁻⁶ Joule
∴ Stored energy is 5.142×10⁻⁶ Joule
Charge
Q = CV
⇒Q = 0.14×10⁻⁶×8.57
⇒Q = 1.2×10⁻⁶ C
∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb
The missing diagram is in the attachments.
Answer: X: positive Y: positive
Explanation: Electric field is a vector quantity, which means it can be represented by a vector arrow: the arrow points in the direction of electric field and its length represents the magnitude at a given location. There are another representation of the electric field called electric field lines, <u>in which the line points away from a positively charged source and towards a negatively charged source</u>. This occurs because it follows a pattern, where the lines points in the direction that a positive test charge would have if it is accelerating on the line.
Analyzing the diagram, it can be observed that the lines are pointing away from both of the charged objects. Therefore, both X and Y are <u>positively charged</u>.